EXTRA EXERCISE FROM JOHN BIRDS BOOK

334 Engineering Mathematics

(d) The amount of money spent on food can only Trends in ungrouped data over equal periods of time
be expressed correct to the nearest pence, the can be presented diagrammatically by a percentage
amount being counted. Hence, these data are component bar chart. In such a chart, equally spaced
discrete. rectangles of any width, but whose height corresponds
to 100%, are constructed. The rectangles are then sub-
Now try the following exercise divided into values corresponding to the percentage
relative frequencies of the members (see Problem 5).
Exercise 134 Further problems on discrete
and continuous data A pie diagram is used to show diagrammatically the
parts making up the whole. In a pie diagram, the area
In Problems 1 and 2, state whether data relating to of a circle represents the whole, and the areas of the
the topics given are discrete or continuous. sectors of the circle are made proportional to the parts
which make up the whole (see Problem 6).

1. (a) The amount of petrol produced daily, for Problem 2. The number of television sets
(b) each of 31 days, by a refinery. repaired in a workshop by a technician in six,
(c) The amount of coal produced daily by one-month periods is as shown below. Present
(d) each of 15 miners. these data as a pictogram.
The number of bottles of milk delivered
daily by each of 20 milkmen. Month January February March
The size of 10 samples of rivets produced 11 6 15
by a machine. Number
repaired
(a) continuous (b) continuous
(c) discrete (d) continuous Month April May June

2. (a) The number of people visiting an exhibi- Number
(b) tion on each of 5 days.
(c) The time taken by each of 12 athletes to repaired 9 13 8
(d) run 100 metres.
Section 7 The value of stamps sold in a day by each Each symbol shown in Fig. 37.1 represents two televi-
of 20 post offices.
The number of defective items produced sion sets repaired. Thus, in January, 5 1 symbols are used
in each of 10 one-hour periods by a 2
machine. to represents the 11 sets repaired, in February, 3 symbols

(a) discrete (b) continuous are used to represent the 6 sets repaired, and so on.
(c) discrete (d) discrete
Month Number of TV sets repaired 2 sets
January
37.2 Presentation of ungrouped data February
March
Ungrouped data can be presented diagrammatically in April
several ways and these include: May
June
(a) pictograms, in which pictorial symbols are used
to represent quantities (see Problem 2), Figure 37.1

(b) horizontal bar charts, having data represented by Problem 3. The distance in miles travelled by
equally spaced horizontal rectangles (see Problem four salesmen in a week are as shown below.
3), and
Salesmen PQR S
(c) vertical bar charts, in which data are repre-
sented by equally spaced vertical rectangles (see Distance
Problem 4). travelled (miles) 413 264 597 143

Use a horizontal bar chart to represent these data
diagrammatically

Presentation of statistical data 335

Equally spaced horizontal rectangles of any width, but percentage component bar charts to present these
whose length is proportional to the distance travelled, data.
are used. Thus, the length of the rectangle for salesman
P is proportional to 413 miles, and so on. The horizontal Year 1 Year 2 Year 3
bar chart depicting these data is shown in Fig. 37.2.

4-roomed bungalows 24 17 7

Salesmen S 5-roomed bungalows 38 71 118
R
Q 4-roomed houses 44 50 53
P
5-roomed houses 64 82 147
0 100 200 300 400 500 600
Distance travelled, miles 6-roomed houses 30 30 25

Figure 37.2

Problem 4. The number of issues of tools or A table of percentage relative frequency values, correct Section 7
materials from a store in a factory is observed for to the nearest 1%, is the first requirement. Since,
seven, one-hour periods in a day, and the results of
the survey are as follows: percentage relative frequency
= frequency of member × 100
Period 1 2 3 4 5 6 7 total frequency

Number then for 4-roomed bungalows in year 1:
of issues 34 17 9 5 27 13 6
percentage relative frequency
Present these data on a vertical bar chart. = 24 × 100 = 12%
24 + 38 + 44 + 64 + 30
In a vertical bar chart, equally spaced vertical rectangles
of any width, but whose height is proportional to the The percentage relative frequencies of the other types of
quantity being represented, are used. Thus the height dwellings for each of the three years are similarly cal-
of the rectangle for period 1 is proportional to 34 units, culated and the results are as shown in the table below.
and so on. The vertical bar chart depicting these data is
shown in Fig. 37.3. Year 1 Year 2 Year 3

4-roomed bungalows 12% 7% 2%

Number of issues 40 5-roomed bungalows 19% 28% 34%
30
20 4-roomed houses 22% 20% 15%
10
5-roomed houses 32% 33% 42%

1234567 6-roomed houses 15% 12% 7%
Periods

Figure 37.3 The percentage component bar chart is produced by con-
structing three equally spaced rectangles of any width,
Problem 5. The number of various types of corresponding to the three years. The heights of the rect-
dwellings sold by a company annually over a angles correspond to 100% relative frequency, and are
three-year period are as shown below. Draw subdivided into the values in the table of percentages
shown above. A key is used (different types of shading
or different colour schemes) to indicate corresponding

336 Engineering Mathematics

percentage values in the rows of the table of percent- Research and
ages. The percentage component bar chart is shown in development
Fig. 37.4.

Labour

100 Percentage relative frequency Key 126° 72° 361°8°Materials
90 6-roomed houses
80 5-roomed houses Overheads 108°
70 4-roomed houses
60 5-roomed bungalows Profit
50 4-roomed bungalows
40 2 l p 1.8°
30 Year 3
20 Figure 37.5
10
(b) Using the data presented in Fig. 37.4,
1 comment on the housing trends over the
three-year period.
Figure 37.4
(c) Determine the profit made by selling 700 units
of the product shown in Fig. 37.5.

Problem 6. The retail price of a product costing (a) By measuring the length of rectangle P the mileage
£2 is made up as follows: materials 10 p, labour covered by salesman P is equivalent to 413
20 p, research and development 40 p, overheads miles. Hence salesman P receives a travelling
70 p, profit 60 p. Present these data on a pie diagram allowance of

£413 × 37
i.e. £152.81

100

Section 7 A circle of any radius is drawn, and the area of the circle Similarly, for salesman Q, the miles travelled are
represents the whole, which in this case is £2. The circle 264 and this allowance is
is subdivided into sectors so that the areas of the sectors
are proportional to the parts, i.e. the parts which make £264 × 37
up the total retail price. For the area of a sector to be i.e. £97.68
proportional to a part, the angle at the centre of the
circle must be proportional to that part. The whole, £2 100
or 200 p, corresponds to 360◦. Therefore,
Salesman R travels 597 miles and he receives
10 p corresponds to 360 × 10 degrees, i.e. 18◦ £597 × 37
200 i.e. £220.89
100
20 p corresponds to 360 × 20 degrees, i.e. 36◦
200 Finally, salesman S receives

and so on, giving the angles at the centre of the circle £143 × 37
for the parts of the retail price as: 18◦, 36◦, 72◦, 126◦ i.e. £52.91
and 108◦, respectively.
The pie diagram is shown in Fig. 37.5. 100

Problem 7. (b) An analysis of Fig. 37.4 shows that 5-roomed bun-

(a) Using the data given in Fig. 37.2 only, galows and 5-roomed houses are becoming more
calculate the amount of money paid to each
salesman for travelling expenses, if they are popular, the greatest change in the three years
paid an allowance of 37 p per mile.
being a 15% increase in the sales of 5-roomed

bungalows.

(c) Since 1.8◦ corresponds to 1 p and the profit occu-

pies 108◦ of the pie diagram, then the profit per

unit is 108 × 1 , that is, 60 p

1.8

The profit when selling 700 units of the product is
700 × 60

£ , that is, £420
100

Presentation of statistical data 337

Now try the following exercise

4. Present the data given in Problem 2 above on

Exercise 135 Further problems on a horizontal⎡bar chart. ⎤
presentation of ungrouped 5 equally spaced
data ⎢⎢⎢⎣⎢ l1he5onr8gi0zth,os2n1taa9rle0r,pe1cro8tap4no0gr,lt2eios3n,8aw5lhatoonsde⎥⎥⎥⎥⎦

1. The number of vehicles passing a stationary 1280 units, respectively.
observer on a road in six ten-minute intervals
is as shown. Draw a pictogram to represent 5. For the data given in Problem 1 above,
these data. construct a vertical bar chart.
⎡⎤
Period of 123456 6 equally spaced vertical
Time ⎢⎣⎢⎢⎢ ra6er2ec,tpa6nr8og,pl4eo9sr,taiwonndhao4ls1teouh3ne5iitg,sh,4t4s,⎥⎥⎥⎦⎥
respectively.
Number of
Vehicles 35 44 62 68 49 41

⎡⎤ 6. Depict the data given in Problem 2 above on
If one symbol is used to
a vertical b⎡ar chart. ⎤
⎢⎢⎢⎣⎢⎢⎢ wgrneieovparerrkeseiss3net.gn55t,cv1o4e0r.h5rvei,cce6lthe,its7co,l,et5hse,and⎥⎥⎥⎥⎥⎥⎦ 5 equally spaced vertical
4 symbols respectively. ⎢⎣⎢⎢⎢ 1pre8rco4tpa0on,rg2tli3eo8sn5,awlatnhodo1s1e528h80e0,igu2hn1ti9st0sa,,re⎦⎥⎥⎥⎥

2. The number of components produced by a respectively.
factory in a week is as shown below:
7. A factory produces three different types of
Day Mon Tues Wed components. The percentages of each of Section 7
these components produced for three, one-
Number of month periods are as shown below. Show
Components 1580 2190 1840 this information on percentage component
bar charts and comment on the changing trend
Day Thurs Fri in the percentages of the types of component
2385 1280 produced.
Number of
Components Month 123

Show these data on a pictogram. Component P 20 35 40
⎡⎤
If one symbol represents Component Q 45 40 35
⎢⎢⎢⎢⎢⎢⎣ 2Mc1o00or00rneccc8oot,mmtToppuootehnnseee1nnn1ttess,a,Wgrweivesodetrsk9:i,ng⎥⎥⎥⎦⎥⎥⎥
Thurs 12 and Fri 6.5 Component R 35 25 25

3. For the data given in Problem 1 above, draw ⎡⎤
Three rectangles of equal
a horizonta⎡l bar chart. ⎤
6 equally spaced horizontal ⎢⎢⎢⎢⎢⎢⎣ cpbhoeeyrilgu2cemh0nt%,ntassaguatebbtsdhoisvevheieod.xwePpdneiinninnscertthehaeeses⎥⎥⎥⎥⎦⎥⎥
⎢⎢⎣ prercotpaonrgtlieosn,awl thoo3se5,le4n4g, t6h2s,are⎥⎥⎦ of Q and R

68, 49 and 41, respectively. 8. A company has five distribution centres and
the mass of goods in tonnes sent to each

338 Engineering Mathematics

centre during four, one-week periods, is as 11. (a) With reference to Fig. 37.5, determine
shown. the amount spent on labour and materials
to produce 1650 units of the product.
Week 1234
(b) If in year 2 of Fig. 37.4, 1% corresponds
Centre A 147 160 174 158 to 2.5 dwellings, how many bungalows
are sold in that year.
Centre B 54 63 77 69 [(a) £495, (b) 88]

Centre C 283 251 237 211 12. (a) If the company sell 23 500 units per
(b) annum of the product depicted in
Centre D 97 104 117 144 Fig. 37.5, determine the cost of their
overheads per annum.
Centre E 224 218 203 194 If 1% of the dwellings represented in
year 1 of Fig. 37.4 corresponds to 2
Use a percentage component bar chart to dwellings, find the total number of
houses sold in that year.
present these data and comment on any
[(a) £16 450, (b) 138]
trends. ⎡
Four rectangles of equal ⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎢ irch29w12neh%983eedcai%%%ergu,nekhc2,,gawtt1721esisoee%47:,ien%%n1so,ku8foc1,b%w2fea92d:nbae5%,i2tvobe%7r0,kiuoe%d%2tsuw3e5t,7Ad:,e%3%828eaa5%2%.ksn%i%Lndf4,i,oin:,3DtB1lt221Cll,2oe0%0aa%,wn%%a,ds,n,,:a⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎥ 37.3 Presentation of grouped data

Section 7 reduction of about 3% in E. When the number of members in a set is small, say ten or
less, the data can be represented diagrammatically with-
9. The employees in a company can be split out further analysis, by means of pictograms, bar charts,
into the following categories: managerial 3, percentage components bar charts or pie diagrams (as
supervisory 9, craftsmen 21, semi-skilled 67, shown in Section 37.2).
others 44. Shown these data on a pie diagram.
⎡⎤ For sets having more than ten members, those mem-
A circle of any radius, bers having similar values are grouped together in
⎢⎢⎢⎣⎢ sh5ua2bv.5din◦iv,gi1da6ne7dg.l5ien◦staoonfsde7c1.5t1o◦0r,◦s2,2.5◦,⎥⎥⎥⎥⎦ classes to form a frequency distribution. To assist in
respectively. accurately counting members in the various classes, a
tally diagram is used (see Problems 8 and 12).
10. The way in which an apprentice spent his
time over a one-month period is a follows: A frequency distribution is merely a table show-
ing classes and their corresponding frequencies (see
drawing office 44 hours, production Problems 8 and 12).
64 hours, training 12 hours, at college
28 hours. The new set of values obtained by forming a fre-
quency distribution is called grouped data.

The terms used in connection with grouped data are
shown in Fig. 37.6(a). The size or range of a class
is given by the upper class boundary value minus
the lower class boundary value, and in Fig. 37.6 is
7.65 – 7.35, i.e. 0.30. The class interval for the class
shown in Fig. 37.6(b) is 7.4 to 7.6 and the class mid-point
value is given by:

Use a pie diagram to depict this information. upper class + lower class
boundary value boundary value
⎡⎤
A circle of any radius, 2

⎣⎢⎢ subdivided into sectors ⎦⎥⎥ and in Fig. 37.6 is 7.65 + 7.35 , i.e. 7.5
having angles of 107◦,
2
156◦, 29◦ and 68◦, respectively.
One of the principal ways of presenting grouped data

diagrammatically is by using a histogram, in which

Presentation of statistical data 339

(a) Class interval The range of the data is the value obtained by tak-
ing the value of the smallest member from that of the
Lower Class Upper largest member. Inspection of the set of data shows
that, range = 89 − 71 = 18. The size of each class is
class mid-point class given approximately by range divided by the number of
classes. Since 7 classes are required, the size of each
boundary boundary class is 18/7, that is, approximately 3. To achieve seven
equal classes spanning a range of values from 71 to 89,
to 7.3 7.4 to 7.6 7.7 to the class intervals are selected as: 70–72, 73–75, and
(b) 7.5 7.65 so on.

7.35 To assist with accurately determining the number in
each class, a tally diagram is produced, as shown in
Figure 37.6 Table 37.1(a). This is obtained by listing the classes
in the left-hand column, and then inspecting each of the
the areas of vertical, adjacent rectangles are made pro- 40 members of the set in turn and allocating them to the
portional to frequencies of the classes (see Problem 9). appropriate classes by putting ‘1s’ in the appropriate
When class intervals are equal, the heights of the rect- rows. Every fifth ‘1’ allocated to a particular row is
angles of a histogram are equal to the frequencies of the shown as an oblique line crossing the four previous ‘1s’,
classes. For histograms having unequal class intervals, to help with final counting.
the area must be proportional to the frequency. Hence,
if the class interval of class A is twice the class inter- Table 37.1(a)
val of class B, then for equal frequencies, the height
of the rectangle representing A is half that of B (see Class Tally
Problem 11).
70–72 1
Another method of presenting grouped data diagram-
matically is by using a frequency polygon, which is the 73–75 11 Section 7
graph produced by plotting frequency against class mid- 76–78 $11$1$1 11
point values and joining the coordinates with straight 79–81 $11$1$1 $11$1$1 11
lines (see Problem 12). 82–84 $11$1$1 1111
85–87 $11$1$1 1
A cumulative frequency distribution is a table
showing the cumulative frequency for each value of 88–90 111
upper class boundary. The cumulative frequency for a
particular value of upper class boundary is obtained by Table 37.1(b)
adding the frequency of the class to the sum of the pre-
vious frequencies. A cumulative frequency distribution Class Class mid-point Frequency
is formed in Problem 13. 1
2
The curve obtained by joining the co-ordinates of 7
cumulative frequency (vertically) against upper class
boundary (horizontally) is called an ogive or a cumu- 12
lative frequency distribution curve (see Problem 13). 9
6
70–72 71 3

Problem 8. The data given below refer to the gain 73–75 74
of each of a batch of 40 transistors, expressed correct
to the nearest whole number. Form a frequency 76–78 77
distribution for these data having seven classes
79–81 80
81 83 87 74 76 89 82 84
86 76 77 71 86 85 87 88 82–84 83
84 81 80 81 73 89 82 79
81 79 78 80 85 77 84 78 85–87 86
83 79 80 83 82 79 80 77
88–90 89

340 Engineering Mathematics

A frequency distribution for the data is shown in Table 37.2 Frequency
Table 37.1(b) and lists classes and their correspond- Class 2
ing frequencies, obtained from the tally diagram. (Class 20–40 6
mid-point values are also shown in the table, since they 50–70 12
are used for constructing the histogram for these data 80–90 14
(see Problem 9)). 4
100–110 2
Problem 9. Construct a histogram for the data 120–140
given in Table 37.1(b) 150–170

Section 7 The histogram is shown in Fig. 37.7. The width of that there are a much smaller number of extreme val-
Frequencythe rectangles correspond to the upper class bound- ues ranging from £30 to £170. If equal class intervals
ary values minus the lower class boundary values and are selected, the frequency distribution obtained does
the heights of the rectangles correspond to the class not give as much information as one with unequal class
frequencies. The easiest way to draw a histogram is intervals. Since the majority of members are between
to mark the class mid-point values on the horizontal £80 and £100, the class intervals in this range are
scale and draw the rectangles symmetrically about the selected to be smaller than those outside of this range.
appropriate class mid-point values and touching one There is no unique solution and one possible solution is
another. shown in Table 37.2.

16 Problem 11. Draw a histogram for the data given
14 in Table 37.2
12
10 When dealing with unequal class intervals, the his-
togram must be drawn so that the areas, (and not the
8 heights), of the rectangles are proportional to the fre-
6 quencies of the classes. The data given are shown in
4 columns 1 and 2 of Table 37.3. Columns 3 and 4 give
2 the upper and lower class boundaries, respectively. In
column 5, the class ranges (i.e. upper class bound-
71 74 77 80 83 86 89 ary minus lower class boundary values) are listed. The
Class mid-point values heights of the rectangles are proportional to the ratio
frequency
Figure 37.7
, as shown in column 6. The histogram is
Problem 10. The amount of money earned weekly class range
by 40 people working part-time in a factory, correct shown in Fig. 37.8.
to the nearest £10, is shown below. Form a frequency
distribution having 6 classes for these data. Problem 12. The masses of 50 ingots in
kilograms are measured correct to the nearest 0.1 kg
80 90 70 110 90 160 110 80 and the results are as shown below. Produce a
140 30 90 50 100 110 60 100 frequency distribution having about 7 classes for
80 90 110 80 100 90 120 70
130 170 80 120 100 110 40 110
50 100 110 90 100 70 110 80

Inspection of the set given shows that the majority of
the members of the set lie between £80 and £110 and

Presentation of statistical data 341

Table 37.3 2 3 4 5 6
1 Frequency Upper class boundary Lower class boundary Class range Height of rectangle

Class 2 45 15 30 2=1
30 15
20–40

50–70 6 75 45 30 6 = 3
80–90 12 95 30 15
100–110 14 115
120–140 145 75 20 12 = 9
150–170 4 175 20 15
2
95 20 14 = 10 1
2

20 15

115 30 4 = 2
30 15

145 30 2 = 1
30 15

12/15 Frequency per unit Since about seven classes are required, the size of each
10/15 class range class is 2.0/7, that is approximately 0.3, and thus the
class limits are selected as 7.1 to 7.3, 7.4 to 7.6, 7.7 to
8/15 Section 77.9, and so on.
6/15 30 60 85 105 130 160
4/15 Class mid-point values The class mid-point for the 7.1 to 7.3 class is
2/15 7.35 + 7.05

Figure 37.8 , i.e. 7.2, for the 7.4 to 7.6 class is
2
these data and then present the grouped data as (a) a 7.65 + 7.35
frequency polygon and (b) histogram.
, i.e. 7.5, and so on.
8.0 8.6 8.2 7.5 8.0 9.1 8.5 7.6 8.2 7.8 2
8.3 7.1 8.1 8.3 8.7 7.8 8.7 8.5 8.4 8.5 To assist with accurately determining the number
7.7 8.4 7.9 8.8 7.2 8.1 7.8 8.2 7.7 7.5 in each class, a tally diagram is produced as shown
8.1 7.4 8.8 8.0 8.4 8.5 8.1 7.3 9.0 8.6 in Table 37.4. This is obtained by listing the classes
7.4 8.2 8.4 7.7 8.3 8.2 7.9 8.5 7.9 8.0 in the left-hand column and then inspecting each of
the 50 members of the set of data in turn and allocat-
The range of the data is the member having the largest ing it to the appropriate class by putting a ‘1’ in the
value minus the member having the smallest value. appropriate row. Each fifth ‘1’ allocated to a particu-
Inspection of the set of data shows that: lar row is marked as an oblique line to help with final
counting.
range = 9.1 − 7.1 = 2.0 A frequency distribution for the data is shown in
Table 37.5 and lists classes and their corresponding fre-
The size of each class is given approximately by quencies. Class mid-points are also shown in this table,
range since they are used when constructing the frequency
polygon and histogram.
number of classes A frequency polygon is shown in Fig. 37.9, the
co-ordinates corresponding to the class mid-point/
frequency values, given in Table 37.5. The co-ordinates
are joined by straight lines and the polygon is ‘anchored-
down’ at each end by joining to the next class mid-point
value and zero frequency.

342 Engineering Mathematics

Table 37.4 Tally 14
Class 12
10 Histogram

8
6
4
2
0

7.2 7.5 7.8 8.1 8.4 8.7 9.0
7.1 to 7.3 111 Frequency
7.4 to 7.6 $11$1$1 7.35
7.7 to 7.9 $11$1$1 1111 7.65
8.0 to 8.2 $11$1$1 $11$1$1 1111 7.95
8.3 to 8.5 $11$1$1 $11$1$1 1 8.25
8.6 to 8.8 $11$1$1 1 8.55
8.85
9.15

Class mid-point values

Figure 37.10

8.9 to 9.1 11 value — lower class boundary value) and height cor-
responding to the class frequency. The easiest way to
Table 37.5 Class mid-point Frequency draw a histogram is to mark class mid-point values on
Class the horizontal scale and to draw the rectangles symmet-
rically about the appropriate class mid-point values and
touching one another. A histogram for the data given in
Table 37.5 is shown in Fig. 37.10.

7.1 to 7.3 7.2 3

7.4 to 7.6 7.5 5 Problem 13. The frequency distribution for the
masses in kilograms of 50 ingots is:
7.5 to 7.9 7.8 9
7.1 to 7.3 3, 7.4 to 7.6 5, 7.7 to 7.9 9,
8.0 to 8.2 8.1 14 8.0 to 8.2 14, 8.3 to 8.5 11, 8.6 to 8.8, 6,
8.9 to 9.1 2,
Section 7 8.3 to 8.5 8.4 11
Form a cumulative frequency distribution for these
8.6 to 8.8 8.7 6 data and draw the corresponding ogive

8.9 to 9.1 9.0 2 A cumulative frequency distribution is a table giv-
ing values of cumulative frequency for the values of
14Frequency upper class boundaries, and is shown in Table 37.6.
12 Frequency polygon Columns 1 and 2 show the classes and their frequen-
10 cies. Column 3 lists the upper class boundary values
for the classes given in column 1. Column 4 gives
8 the cumulative frequency values for all frequencies less
6 than the upper class boundary values given in column
4 3. Thus, for example, for the 7.7 to 7.9 class shown
2 in row 3, the cumulative frequency value is the sum
0 of all frequencies having values of less than 7.95, i.e.
3 + 5 + 9 = 17, and so on. The ogive for the cumula-
7.2 7.5 7.8 8.1 8.4 8.7 9.0 tive frequency distribution given in Table 37.6 is shown
Class mid-point values in Fig. 37.11. The co-ordinates corresponding to each
upper class boundary/cumulative frequency value are
Figure 37.9 plotted and the co-ordinates are joined by straight lines
(— not the best curve drawn through the co-ordinates
A histogram is shown in Fig. 37.10, the width of as in experimental work). The ogive is ‘anchored’ at its
a rectangle corresponding to (upper class boundary start by adding the co-ordinate (7.05, 0).

Presentation of statistical data 343

Table 37.6 2 3 4 39.7 40.4 39.9 40.1 39.9
Frequency Upper Class Cumulative
1 frequency 39.5 40.0 39.8 39.5 39.9
Class boundary
Less than 3 40.1 40.0 39.7 40.4 39.3
8
7.35 17 40.7 39.9 40.2 39.9 40.0
7.65 31
7.95 42 40.1 39.7 40.5 40.5 39.9
8.25 48
8.55 50 40.8 40.0 40.2 40.0 39.9
8.85
9.15 39.8 39.7 39.5 40.1 40.2

7.1–7.3 3 40.6 40.1 39.7 40.2 40.3
7.4–7.6 5 ⎡⎤
7.7–7.9 9 There is no unique solution,
8.0–8.2 14 ⎢⎢⎢⎢⎢⎢⎣334909...371–––334990b...u842t 1577;;;⎦⎥⎥⎥⎥⎥⎥
8.3–8.5 11 one solution is:
8.6–8.8 6 1; 39.5–39.6
8.9–9.1 2 9; 39.9–40.0
15; 40.3–40.4

40.5–40.6 4; 40.7–40.8 2

2. Draw a histogram for the frequency distribu-

tion given in the solution of Problem 1.
⎡⎤
Rectangles, touching one another,
⎣⎢⎢ ⎥⎥⎦
having mid-points of 39.35,
39.55, 39.75, 39.95, … and

heights of 1, 5, 9, 17, …

50 3. The information given below refers to the value
of resistance in ohms of a batch of 48 resistors
40 of similar value. Form a frequency distribution
for the data, having about 6 classes and draw a
30 frequency polygon and histogram to represent
these data diagrammatically.
Cumulative frequency20

Section 710
21.0 22.4 22.8 21.5 22.6 21.1 21.6 22.3
7.05 7.35 7.65 7.95 8.25 8.55 8.85 9.15 22.9 20.5 21.8 22.2 21.0 21.7 22.5 20.7
Upper class boundary values in kilograms 23.2 22.9 21.7 21.4 22.1 22.2 22.3 21.3
22.1 21.8 22.0 22.7 21.7 21.9 21.1 22.6
Figure 37.11 21.4 22.4 22.3 20.9 22.8 21.2 22.7 21.6
22.2 21.6 21.3 22.1 21.5 22.0 23.4 21.2
Now try the following exercise
⎡⎤
Exercise 136 Further problems on There is no unique solution,
presentation of grouped data ⎢⎢⎣⎢⎢ ⎥⎥⎥⎦⎥
but one solution is: 10;
1. The mass in kilograms, correct to the nearest 20.5–20.9 3; 21.0–21.4 13;
one-tenth of a kilogram, of 60 bars of metal 21.5–21.9
are as shown. Form a frequency distribution 11; 22.0–22.4
of about 8 classes for these data.
22.5–22.9 9; 23.0–23.4 2
39.8 40.3 40.6 40.0 39.6
39.6 40.2 40.3 40.4 39.8 4. The time taken in hours to the failure of 50
40.2 40.3 39.9 39.9 40.0 specimens of a metal subjected to fatigue fail-
40.1 40.0 40.1 40.1 40.2 ure tests are as shown. Form a frequency dis-
tribution, having about 8 classes and unequal
class intervals, for these data.

28 22 23 20 12 24 37 28 21 25
21 14 30 23 27 13 23 7 26 19
24 22 26 3 21 24 28 40 27 24
20 25 23 26 47 21 29 26 22 33
27 9 13 35 20 16 20 25 18 22

344 Engineering Mathematics

⎡ There is no unique solution, ⎤ (b) Draw a histogram depicting the data.

⎢⎣⎢ but one solution is: 1–10 3; ⎥⎥⎦ (c) Form a cumulative frequency distribu-
11–19 7; 20–22 12; 23–25 11;
tion.

26–28 10; 29–38 5; 39–48 2 (d) Draw an ogive for the the data.
⎡⎤
(a) There is unique solution,
5. Form a cumulative frequency distribution and ⎢⎢⎢⎢⎢⎢⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
hence draw the ogive for the frequency distri- (b) but one solution is: 10;
bution given in the solution to Problem 3. (c) 2.05–2.09 3; 2.10–2.14 13;
(d) 2.15–2.19 11; 2.20–2.24
20.95 3; 21.45 13; 21.95 24; 2.25–2.29 9; 2.30–2.34 2
22.45 37; 22.95 46; 23.45 48 Rectangles, touching one
another, having mid-points of 24;
6. Draw a histogram for the frequency distribu- 2.07, 2.12 … and heights of 48
3, 10, …
tion g⎡iven in the solution to Problem 4. ⎤ Using the frequency
Rectangles, touching one another, distribution given in the
⎢⎣⎢⎢⎢⎢⎢ ⎥⎥⎦⎥⎥⎥⎥ solution to part (a) gives:
having mid-points of 5.5, 15, 2.095 3; 2.145 13; 2.195
21, 24, 27, 33.5 and 43.5. The 2.245 37; 2.295 46; 2.345
heights of the rectangles (frequency A graph of cumulative
per unit class range) are 0.3, frequency against upper
class boundary having
0.78, 4. 4.67, 2.33, 0.5 and 0.2 the coordinates given

7. The frequency distribution for a batch of in part (c).
50 capacitors of similar value, measured in
Section 7 microfarads, is:

10.5–10.9 2, 11.0–11.4 7,
11.5–11.9 10, 12.0–12.4 12,
12.5–12.9 11, 13.0–13.4 8

Form a cumulative frequency distribution for
these data.

(10.95 2), (11.45 9), (11.95 11),
(12.45 31), (12.95 42), (13.45 50)

8. Draw an ogive for the data given in the
solution of Problem 7.

9. The diameter in millimetres of a reel of wire
is measured in 48 places and the results are as
shown.

2.10 2.29 2.32 2.21 2.14 2.22
2.28 2.18 2.17 2.20 2.23 2.13
2.26 2.10 2.21 2.17 2.28 2.15
2.16 2.25 2.23 2.11 2.27 2.34
2.24 2.05 2.29 2.18 2.24 2.16
2.15 2.22 2.14 2.27 2.09 2.21
2.11 2.17 2.22 2.19 2.12 2.20
2.23 2.07 2.13 2.26 2.16 2.12

(a) Form a frequency distribution of
diameters having about 6 classes.

Chapter 38

Measures of central tendency
and dispersion

38.1 Measures of central tendency set: {7, 5, 74, 10} has a mean value of 24, which is not
really representative of any of the values of the members
A single value, which is representative of a set of values, of the set. The median value is obtained by:
may be used to give an indication of the general size of
the members in a set, the word ‘average’ often being (a) ranking the set in ascending order of magnitude,
used to indicate the single value. and

The statistical term used for ‘average’ is the arith- (b) selecting the value of the middle member for sets
metic mean or just the mean. Other measures of central containing an odd number of members, or finding
tendency may be used and these include the median and the value of the mean of the two middle members
the modal values. for sets containing an even number of members.

38.2 Mean, median and mode for For example, the set: {7, 5, 74, 10} is ranked as
discrete data {5, 7, 10, 74}, and since it contains an even num-
ber of members (four in this case), the mean of
Mean 7 and 10 is taken, giving a median value of 8.5.
Similarly, the set: {3, 81, 15, 7, 14} is ranked as
The arithmetic mean value is found by adding together {3, 7, 14, 15, 81} and the median value is the value
the values of the members of a set and dividing by the of the middle member, i.e. 14.
number of members in the set. Thus, the mean of the
set of numbers: {4, 5, 6, 9} is: Mode

4+5+6+9 The modal value, or mode, is the most commonly
i.e. 6 occurring value in a set. If two values occur with
the same frequency, the set is ‘bi-modal’. The set:
4 {5, 6, 8, 2, 5, 4, 6, 5, 3} has a modal value of 5, since
In general, the mean of the set: {x1, x2, x3, …, xn} is the member having a value of 5 occurs three times.

x = x1 + x2 + x3 + · · · + xn , written as x Problem 1. Determine the mean, median and
nn mode for the set:

where is the Greek letter ‘sigma’ and means ‘the sum {2, 3, 7, 5, 5, 13, 1, 7, 4, 8, 3, 4, 3}
of’, and x (called x-bar) is used to signify a mean value.
The mean value is obtained by adding together the val-
Median ues of the members of the set and dividing by the number
of members in the set.
The median value often gives a better indication of
the general size of a set containing extreme values. The

346 Engineering Mathematics

Thus, mean value, 2. {26, 31, 21, 29, 32, 26, 25, 28}
[mean 27.25, median 27, mode 26]
2 + 3 + 7 + 5 + 5 + 13 + 1
x = +7 + 4 + 8 + 3 + 4 + 3 = 65 = 5 3. {4.72, 4.71, 4.74, 4.73, 4.72, 4.71, 4.73, 4.72}
[mean 4.7225, median 4.72, mode 4.72]
13 13
4. {73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9}
To obtain the median value the set is ranked, that is, [mean 115.2, median 126.4, no mode]
placed in ascending order of magnitude, and since the
set contains an odd number of members the value of the 38.3 Mean, median and mode for
middle member is the median value. Ranking the set grouped data
gives:
The mean value for a set of grouped data is found
{1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13} by determining the sum of the (frequency × class mid-
point values) and dividing by the sum of the frequencies,
The middle term is the seventh member, i.e. 4, thus the
median value is 4. i.e. mean value x = f1x1 + f2x2 + · · · + fnxn
f1 + f2 + · · · + fn
The modal value is the value of the most commonly
occurring member and is 3, which occurs three times, = (fx)
all other members only occurring once or twice. f

Problem 2. The following set of data refers to the where f is the frequency of the class having a mid-point
amount of money in £s taken by a news vendor for value of x, and so on.
6 days. Determine the mean, median and modal
values of the set: Problem 3. The frequency distribution for the
value of resistance in ohms of 48 resistors is as
{27.90, 34.70, 54.40, 18.92, 47.60, 39.68} shown. Determine the mean value of resistance.

27.90 + 34.70 + 54.40 20.5–20.9 3, 21.0–21.4 10, 21.5–21.9 11,
22.0–22.4 13, 22.5–22.9 9, 23.0–23.4 2
Mean value = +18.92 + 47.60 + 39.68 = £37.20
The class mid-point/frequency values are:
6
20.7 3, 21.2 10, 21.7 11, 22.2 13,
Section 7 The ranked set is: 22.7 9 and 23.2 2

{18.92, 27.90, 34.70, 39.68, 47.60, 54.40} For grouped data, the mean value is given by:
x = (fx)
Since the set has an even number of members, the mean f
of the middle two members is taken to give the median
value, i.e. where f is the class frequency and x is the class
mid-point value. Hence mean value,
median value = 34.70 + 39.68 = £37.19
2 (3 × 20.7) + (10 × 21.2) + (11 × 21.7)
+(13 × 22.2) + (9 × 22.7) + (2 × 23.2)
Since no two members have the same value, this set has x=
no mode.
48
Now try the following exercise = 1052.1 = 21.919 . . .

Exercise 137 Further problems on mean, 48
median and mode for i.e. the mean value is 21.9 ohms, correct to 3 significant
discrete data figures.

In Problems 1 to 4, determine the mean, median
and modal values for the sets given.

1. {3, 8, 10, 7, 5, 14, 2, 9, 8}
[mean 7.33, median 8, mode 8]

Measures of central tendency and dispersion 347

Histogram rectangles are shown circled on the histogram giving a Section 7
The mean, median and modal values for grouped data total area of 100 square units. The positions, m, of the
may be determined from a histogram. In a histogram, centroids of the individual rectangles are 1, 3, 5, …units
frequency values are represented vertically and variable from YY. Thus
values horizontally. The mean value is given by the value
of the variable corresponding to a vertical line drawn 100 M = (10 × 1) + (16 × 3) + (32 × 5)
through the centroid of the histogram. The median value
is obtained by selecting a variable value such that the + (24 × 7) + (12 × 9) + (6 × 11)
area of the histogram to the left of a vertical line drawn i.e. M = 560 = 5.6 units from YY
through the selected variable value is equal to the area of
the histogram on the right of the line. The modal value 100
is the variable value obtained by dividing the width of Thus the position of the mean with reference to the time
the highest rectangle in the histogram in proportion to scale is 14 + 5.6, i.e. 19.6 minutes.
the heights of the adjacent rectangles. The method of
determining the mean, median and modal values from The median is the value of time corresponding to a
a histogram is shown in Problem 4. vertical line dividing the total area of the histogram into
two equal parts. The total area is 100 square units, hence
Problem 4. The time taken in minutes to the vertical line must be drawn to give 50 units of area on
assemble a device is measured 50 times and the each side. To achieve this with reference to Fig. 38.1,
results are as shown. Draw a histogram depicting rectangle ABFE must be split so that 50 − (10 + 16)
this data and hence determine the mean, median units of area lie on one side and 50 − (24 + 12 + 6) units
and modal values of the distribution. of area lie on the other. This shows that the area of ABFE
14.5–15.5 5, 16.5–17.5 8, 18.5–19.5 16, is split so that 24 units of area lie to the left of the line
20.5–21.5 12, 22.5–23.5 6, 24.5–25.5 3 and 8 units of area lie to the right, i.e. the vertical line
must pass through 19.5 minutes. Thus the median value
The histogram is shown in Fig. 38.1. The mean value of the distribution is 19.5 minutes.
lies at the centroid of the histogram. With reference to
any arbitrary axis, sayYY shown at a time of 14 minutes, The mode is obtained by dividing the line AB, which
the position of the horizontal value of the centroid can be is the height of the highest rectangle, proportionally to
obtained from the relationship AM = (am), where A is the heights of the adjacent rectangles. With reference to
the area of the histogram, M is the horizontal distance of Fig. 38.1, this is done by joining AC and BD and drawing
the centroid from the axis YY, a is the area of a rectangle a vertical line through the point of intersection of these
of the histogram and m is the distance of the centroid two lines. This gives the mode of the distribution and
of the rectangle from YY. The areas of the individual is 19.3 minutes.

Figure 38.1 Now try the following exercise

Exercise 138 Further problems on mean,
median and mode for
grouped data

1. 21 bricks have a mean mass of 24.2 kg, and
29 similar bricks have a mass of 23.6 kg.
Determine the mean mass of the 50 bricks.
[23.85 kg]

2. The frequency distribution given below refers
to the heights in centimetres of 100 people.
Determine the mean value of the distribution,
correct to the nearest millimetre.

150–156 5, 157–163 18, 164–170 20
171–177 27, 178–184 22, 185–191 8

[171.7 cm]

348 Engineering Mathematics

3. The gain of 90 similar transistors is measured (e) divide by the number of members in the set, n,
and the results are as shown. giving
(x1 − x)2 + (x2 − x)2 + (x3 − x)2 + · · ·
83.5–85.5 6, 86.5–88.5 39, 89.5–91.5 27, n
92.5–94.5 15, 95.5–97.5 3
(f) determine the square root of (e).

By drawing a histogram of this frequency dis- The standard deviation is indicated by σ (the Greek letter
tribution, determine the mean, median and small ‘sigma’) and is written mathematically as:
modal values of the distribution.
standard deviation, σ = (x − x)2
[mean 89.5, median 89, mode 88.2] n

4. The diameters, in centimetres, of 60 holes where x is a member of the set, x is the mean value of the
bored in engine castings are measured and
the results are as shown. Draw a histogram set and n is the number of members in the set. The value
depicting these results and hence determine
the mean, median and modal values of the of standard deviation gives an indication of the distance
distribution.
of the members of a set from the mean value. The set:

{1, 4, 7, 10, 13} has a mean value of 7 and a standard

deviation of about 4.2. The set {5, 6, 7, 8, 9} also has

2.011–2.014 7, 2.016–2.019 16, a mean value of 7, but the standard deviation is about
2.021–2.024 23, 2.026–2.029 9,
2.031–2.034 1.4. This shows that the members of the second set are
5
mainly much closer to the mean value than the members

mean 2.02158 cm, median 2.02152 cm, of the first set. The method of determining the standard
mode 2.02167 cm
deviation for a set of discrete data is shown in Problem 5.

Problem 5. Determine the standard deviation
from the mean of the set of numbers:
{5, 6, 8, 4, 10, 3}, correct to 4 significant figures.

Section 7 38.4 Standard deviation The arithmetic mean, x = x
n
(a) Discrete data
5+6+8+4
The standard deviation of a set of data gives an indi- +10 + 3
cation of the amount of dispersion, or the scatter, of
members of the set from the measure of central ten- = =6
dency. Its value is the root-mean-square value of the 6
members of the set and for discrete data is obtained as
follows: Standard deviation, σ = (x − x)2
n
(a) determine the measure of central tendency, usually
the mean value, (occasionally the median or modal The (x − x)2 values are: (5 − 6)2, (6 − 6)2, (8 − 6)2,
values are specified), (4 − 6)2, (10 − 6)2 and (3 − 6)2.

(b) calculate the deviation of each member of the set The sum of the (x − x)2 values,
from the mean, giving
i.e. (x − x)2 = 1 + 0 + 4 + 4 + 16 + 9 = 34
(x1 − x), (x2 − x), (x3 − x), . . . ,
and (x − x)2 = 34 = 5.6˙
(c) determine the squares of these deviations, i.e. n6

(x1 − x)2, (x2 − x)2, (x3 − x)2, . . . , since there are 6 members in the set.
Hence, standard deviation,
(d) find the sum of the squares of the deviations,
that is σ = (x − x2) = 5.6˙ = 2.380
n
(x1 − x)2 + (x2 − x)2 + (x3 − x)2, . . . ,
correct to 4 significant figures

Measures of central tendency and dispersion 349

(b) Grouped data Now try the following exercise
For grouped data, standard deviation
Exercise 139 Further problems on
σ = { f (x − x)2} standard deviation
f
1. Determine the standard deviation from the
where f is the class frequency value, x is the class mid- mean of the set of numbers:
point value and x is the mean value of the grouped data.
The method of determining the standard deviation for a {35, 22, 25, 23, 28, 33, 30}
set of grouped data is shown in Problem 6.
correct to 3 significant figures. [4.60]
Problem 6. The frequency distribution for the
values of resistance in ohms of 48 resistors is as 2. The values of capacitances, in microfarads, of
shown. Calculate the standard deviation from the ten capacitors selected at random from a large
mean of the resistors, correct to 3 significant figures. batch of similar capacitors are:
20.5–20.9 3, 21.0–21.4 10, 21.5–21.9 11,
22.0–22.4 13, 22.5–22.9 9, 23.0–23.4 2 34.3, 25.0, 30.4, 34.6, 29.6, 28.7,
33.4, 32.7, 29.0 and 31.3
The standard deviation for grouped data is given by:
Determine the standard deviation from the
f (x − x)2
σ= mean for these capacitors, correct to 3 signifi-

f cant figures. [2.83 μF]
From Problem 3, the distribution mean value, x = 21.92,
correct to 4 significant figures. 3. The tensile strength in megapascals for 15 Section 7
samples of tin were determined and found
The ‘x-values’ are the class mid-point values, i.e. 20.7, to be:
21.2, 21.7, …,
34.61, 34.57, 34.40, 34.63, 34.63, 34.51,
Thus the (x − x)2 values are (20.7 − 21.92)2, 34.49, 34.61, 34.52, 34.55, 34.58, 34.53,
(21.2 − 21.92)2, (21.7 − 21.92)2, …. 34.44, 34.48 and 34.40

and the f (x − x)2 values are 3(20.7 − 21.92)2, Calculate the mean and standard deviation
10(21.2 − 21.92)2, 11(21.7 − 21.92)2, …. from the mean for these 15 values, correct to
4 significant figures.
The f (x − x)2 values are
4.4652 + 5.1840 + 0.5324 + 1.0192 mean 34.53 MPa, standard
+ 5.4756 + 3.2768 = 19.9532 deviation 0.07474 MPa

f (x − x)2 = 19.9532 = 0.41569 4. Calculate the standard deviation from the mean
f 48
for the mass of the 50 bricks given in Prob-
and standard deviation,
lem 1 of Exercise 138, page 347, correct to 3
{ f (x − x)2}
σ= significant figures. [0.296 kg]

f 5. Determine the standard deviation from the
√
= 0.41569 = 0.645, mean, correct to 4 significant figures, for the

correct to 3 significant figures heights of the 100 people given in Problem 2

of Exercise 138, page 347. [9.394 cm]

6. Calculate the standard deviation from the mean
for the data given in Problem 4 of Exercise 138,
page 348, correct to 3 significant figures.
[0.00544 cm]

350 Engineering Mathematics

38.5 Quartiles, deciles and Figure 38.2
percentiles
corresponding to the first quartile, 12 to 24 correspond-
Other measures of dispersion, which are sometimes
used, are the quartile, decile and percentile values. The ing to the second quartile, 24 to 36 corresponding to the
quartile values of a set of discrete data are obtained by
selecting the values of members that divide the set into third quartile and 36 to 48 corresponding to the fourth
four equal parts. Thus for the set: {2, 3, 4, 5, 5, 7, 9,
11, 13, 14, 17} there are 11 members and the values of quartile of the distribution, i.e. the distribution is divided
the members dividing the set into four equal parts are
4, 7, and 13. These values are signified by Q1, Q2 and into four equal parts. The quartile values are those of the
Q3 and called the first, second and third quartile val-
ues, respectively. It can be seen that the second quartile variable corresponding to cumulative frequency values
value, Q2, is the value of the middle member and hence
is the median value of the set. of 12, 24 and 36, marked Q1, Q2 and Q3 in Fig. 38.2.

For grouped data the ogive may be used to determine These values, correct to the nearest hour, are 37 hours,
the quartile values. In this case, points are selected on the
vertical cumulative frequency values of the ogive, such 43 hours and 48 hours, respectively. The Q2 value is
that they divide the total value of cumulative frequency
into four equal parts. Horizontal lines are drawn from also equal to the median value of the distribution. One
these values to cut the ogive. The values of the variable
corresponding to these cutting points on the ogive give measure of the dispersion of a distribution is called the
the quartile values (see Problem 7).
semi-interquartile range and is given by: (Q3 − Q1)/2,
When a set contains a large number of members, the 1
set can be split into ten parts, each containing an equal and is (48 − 37)/2 in this case, i.e. 5 2 hours.
number of members. These ten parts are then called
Section 7 deciles. For sets containing a very large number of mem- Problem 8. Determine the numbers contained in
bers, the set may be split into one hundred parts, each the (a) 41st to 50th percentile group, and (b) 8th
containing an equal number of members. One of these decile group of the set of numbers shown below:
parts is called a percentile.
14 22 17 21 30 28 37 7 23 32
Problem 7. The frequency distribution given 24 17 20 22 27 19 26 21 15 29
below refers to the overtime worked by a group of
craftsmen during each of 48 working weeks in a The set is ranked, giving:
year.
7 14 15 17 17 19 20 21 21 22
25–29 5, 30–34 4, 35–39 7, 40–44 11,
45–49 12, 50–54 8, 55–59 1 22 23 24 26 27 28 29 30 32 37

Draw an ogive for this data and hence determine the (a) There are 20 numbers in the set, hence the first 10%
quartile values. will be the two numbers 7 and 14, the second 10%
will be 15 and 17, and so on. Thus the 41st to 50th
The cumulative frequency distribution (i.e. upper class percentile group will be the numbers 21 and 22
boundary/cumulative frequency values) is:
(b) The first decile group is obtained by splitting the
29.5 5, 34.5 9, 39.5 16, ranked set into 10 equal groups and selecting the
44.5 27, 49.5 39, 54.5 47, first group, i.e. the numbers 7 and 14. The second
59.5 48 decile group are the numbers 15 and 17, and so on.
Thus the 8th decile group contains the numbers 27
The ogive is formed by plotting these values on a graph, and 28
as shown in Fig. 38.2. The total frequency is divided
into four equal parts, each having a range of 48/4,
i.e. 12. This gives cumulative frequency values of 0 to 12

Now try the following exercise Measures of central tendency and dispersion 351

Exercise 140 Further problems on quartiles, 3. Determine the quartile values and semi-
deciles and percentiles interquartile range for the frequency distri-
bution given in Problem 2 of Exercise 138,
1. The number of working days lost due to acci- page 347.
dents for each of 12 one-monthly periods are Q1 = 164.5 cm, Q2 = 172.5 cm,
as shown. Determine the median and first and Q3 = 179 cm, 7.25 cm
third quartile values for this data.
4. Determine the numbers contained in the 5th
27 37 40 28 23 30 decile group and in the 61st to 70th percentile
35 24 30 32 31 28 groups for the set of numbers:

[30, 25.5, 33.5 days] 40 46 28 32 37 42 50 31 48 45
32 38 27 33 40 35 25 42 38 41
2. The number of faults occurring on a production
line in a nine-week period are as shown below. [37 and 38; 40 and 41]
Determine the median and quartile values for
the data. 5. Determine the numbers in the 6th decile group
and in the 81st to 90th percentile group for the
30 27 25 24 27 set of numbers:
37 31 27 35
43 47 30 25 15 51 17 21 37 33 44 56 40 49 22
[27, 26, 33 faults] 36 44 33 17 35 58 51 35 44 40 31 41 55 50 16

[40, 40, 41; 50, 51, 51]

Section 7

Chapter 39

Probability

39.1 Introduction to probability Dependent event

The probability of something happening is the likeli- A dependent event is one in which the probability of
hood or chance of it happening. Values of probability lie an event happening affects the probability of another
between 0 and 1, where 0 represents an absolute impos- event happening. Let 5 transistors be taken at random
sibility and 1 represents an absolute certainty. The prob- from a batch of 100 transistors for test purposes, and the
ability of an event happening usually lies somewhere probability of there being a defective transistor, p1, be
between these two extreme values and is expressed determined. At some later time, let another 5 transistors
either as a proper or decimal fraction. Examples of be taken at random from the 95 remaining transistors in
probability are: the batch and the probability of there being a defective
transistor, p2, be determined. The value of p2 is differ-
that a length of copper wire 0 ent from p1 since batch size has effectively altered from
has zero resistance at 100◦C 100 to 95, i.e. probability p2 is dependent on probability
1 or 0.1667 p1. Since transistors are drawn, and then another 5 tran-
that a fair, six-sided dice 6 sistors drawn without replacing the first 5, the second
will stop with a 3 upwards random selection is said to be without replacement.
1 or 0.5
that a fair coin will land 2 Independent event
with a head upwards
1 An independent event is one in which the probability
that a length of copper wire of an event happening does not affect the probability
has some resistance at 100◦C of another event happening. If 5 transistors are taken at
random from a batch of transistors and the probability of
If p is the probability of an event happening and q is the a defective transistor p1 is determined and the process
probability of the same event not happening, then the is repeated after the original 5 have been replaced in
total probability is p + q and is equal to unity, since it is the batch to give p2, then p1 is equal to p2. Since the
an absolute certainty that the event either does or does 5 transistors are replaced between draws, the second
not occur, i.e. p + q = 1 selection is said to be with replacement.

Expectation Conditional probability

The expectation, E, of an event happening is defined Conditional probability is concerned with the proba-
bility of say event B occurring, given that event A has
in general terms as the product of the probability p of already taken place. If A and B are independent events,
then the fact that event A has already occurred will not
an event happening and the number of attempts made, affect the probability of event B. If A and B are depen-
dent events, then event A having occurred will effect the
n, i.e. E = pn. probability of event B.

Thus, since the probability of obtaining a 3 upwards

when rolling a fair dice is 1 , the expectation of getting
6
1 2
a 3 upwards on four throws of the dice is 6 × 4, i.e. 3

Thus expectation is the average occurrence of an

event.

Probability 353

39.2 Laws of probability (Check: the total probability should be equal to 1;

The addition law of probability p = 20 and q = 33
,
The addition law of probability is recognized by the 53 53
word ‘or’ joining the probabilities. If pA is the proba-
bility of event A happening and pB is the probability of thus the total probability,
event B happening, the probability of event A or event
B happening is given by pA + pB (provided events A p + q = 20 + 33 = 1
and B are mutually exclusive, i.e. A and B are events 53 53
which cannot occur together). Similarly, the probability
of events A or B or C or . . . N happening is given by hence no obvious error has been made.)

pA + pB + pC + · · · + pN Problem 2. Find the expectation of obtaining a 4
upwards with 3 throws of a fair dice
The multiplication law of probability
Expectation is the average occurrence of an event
The multiplication law of probability is recognized by
the word ‘and’ joining the probabilities. If pA is the and is defined as the probability times the number of
probability of event A happening and pB is the proba-
bility of event B happening, the probability of event A attempts. The probability, p, of obtaining a 4 upwards
and event B happening is given by pA × pB. Similarly,
the probability of events A and B and C and . . . N for one throw of the dice, is 1 .
happening is given by: 6
Also, 3 attempts are made, hence n = 3 and the
pA × pB × pC × · · · × pN
expectation, E, is pn, i.e.
39.3 Worked problems on
probability E = 1 × 3 = 1 or 0.50
6 2

Problem 3. Calculate the probabilities of
selecting at random:

(a) the winning horse in a race in which 10 horses Section 7
are running

(b) the winning horses in both the first and second
races if there are 10 horses in each race

Problem 1. Determine the probabilities of (a) Since only one of the ten horses can win, the
selecting at random (a) a man, and (b) a woman probability of selecting at random the winning
from a crowd containing 20 men and 33 women horse is

number of winners 1
(a) The probability of selecting at random a man, p, is i.e. or 0.10
given by the ratio number of horses 10

number of men (b) The probability of selecting the winning horse
number in crowd 1

20 20 in the first race is . The probability of selecting
i.e. p = 20 + 33 = 53 or 0.3774 10
1
(b) The probability of selecting at random a woman,
q, is given by the ratio the winning horse in the second race is . The
10

probability of selecting the winning horses in the

first and second race is given by the multiplication

law of probability,

number of women i.e. probability = 1 × 1
number in crowd 10 10

i.e. q = 33 = 33 or 0.6226 = 1 or 0.01
20 + 33 53 100

354 Engineering Mathematics

Problem 4. The probability of a component capacitor: (a) both are within the required tolerance
failing in one year due to excessive temperature is values when selecting with replacement, and (b) the
11 first one drawn is below and the second one drawn
is above the required tolerance value, when
, due to excessive vibration is and due to selection is without replacement
20 25
(a) The probability of selecting a capacitor within the
1 73
excessive humidity is . Determine the
required tolerance values is . The first capac-
50 100
probabilities that during a one-year period a
component: (a) fails due to excessive (b) fails due itor drawn is now replaced and a second one is
to excessive vibration or excessive humidity, and drawn from the batch of 100. The probability of
(c) will not fail because of both excessive this capacitor being within the required tolerance
temperature and excessive humidity
73
Let pA be the probability of failure due to excessive values is also .
temperature, then
100
pA = 1 and pA = 19 Thus, the probability of selecting a capacitor
20 20 within the required tolerance values for both the
first and the second draw is:
(where pA is the probability of not failing.)
73 × 73 = 5329 or 0.5329
Let pB be the probability of failure due to excessive 100 100 10 000
vibration, then
(b) The probability of obtaining a capacitor below the
1 24 17
pB = 25 and pB = 25
required tolerance values on the first draw is .
Let pC be the probability of failure due to excessive 100
humidity, then
There are now only 99 capacitors left in the batch,
pC = 1 and pC = 49 since the first capacitor is not replaced. The proba-
50 50 bility of drawing a capacitor above the required tol-

Section 7 (a) The probability of a component failing due to 10
excessive temperature and excessive vibration is erance values on the second draw is , since there
given by:
99
pA × pB = 1 × 1 = 1 or 0.002 are (100 − 73 − 17), i.e. 10 capacitors above the
20 25 500 required tolerance value. Thus, the probability of
randomly selecting a capacitor below the required
(b) The probability of a component failing due to tolerance values and followed by randomly select-
excessive vibration or excessive humidity is: ing a capacitor above the tolerance values is

17 × 10 = 170 = 17 or 0.0172
100 99 9900 990

11 3 Now try the following exercise
pB + pC = 25 + 50 = 50 or 0.06

(c) The probability that a component will not fail due Exercise 141 Further problems on
to excessive temperature and will not fail due to probability
excess humidity is:

19 49 931 1. In a batch of 45 lamps there are 10 faulty
pA × pC = 20 × 50 = 1000 or 0.931
lamps. If one lamp is drawn at random,

Problem 5. A batch of 100 capacitors contains 73 find the probability of it being (a) faulty and
that are within the required tolerance values, 17
which are below the required tolerance values, and (b) satisfactory. ⎡2 ⎤
the remainder are above the required tolerance
values. Determine the probabilities that when ⎢⎣⎢ (a) 9 or 0.2222 ⎥⎥⎦
randomly selecting a capacitor and then a second 7
(b) or 0.7778
9

2. A box of fuses are all of the same shape and Probability 355

size and comprises 23 2 A fuses, 47 5 A fuses 39.4 Further worked problems on
probability
and 69 13 A fuses. Determine the probability
Problem 6. A batch of 40 components contains 5
of selecting at random (a) a 2 A fuse, (b) a 5 A which are defective. A component is drawn at
random from the batch and tested and then a second
fuse and (c) a 13 A fuse. component is drawn. Determine the probability that
⎡ 23 ⎤ neither of the components is defective when drawn
(a) with replacement, and (b) without replacement.
⎢⎢⎣⎢⎢⎢⎢ (a) 139 or 0.1655 ⎥⎥⎥⎥⎥⎦⎥
(b) 47 or 0.3381 (a) With replacement
139
69 The probability that the component selected on the first
(c) or 0.4964 35 7
139
draw is satisfactory is , i.e. . The component is now
3. (a) Find the probability of having a 2 upwards 40 8

when throwing a fair 6-sided dice. (b) Find replaced and a second draw is made. The probability
7
the probability of having a 5 upwards when
that this component is also satisfactory is . Hence, the
throwing a fair 6-sided dice. (c) Determine the 8

probability of having a 2 and then a 5 on two probability that both the first component drawn and the
second component drawn are satisfactory is:
successive throws of a fair 6-sided dice.
7 × 7 = 49 or 0.7656
11 1 8 8 64
(a) (b) (c)
6 6 36 (b) Without replacement

4. Determine the probability that the total score The probability that the first component drawn is sat-
is 8 when two like dice are thrown. 7
5
isfactory is . There are now only 34 satisfactory
36 8

5. The probability of event A happening is 3 and components left in the batch and the batch number is 39. Section 7
5 Hence, the probability of drawing a satisfactory compo-
2
the probability of event B happening is 3 . Cal- 34
nent on the second draw is . Thus the probability that
culate the probabilities of (a) both A and B
39
happening, (b) only event A happening, i.e. the first component drawn and the second component
drawn are satisfactory, i.e. neither is defective, is:
event A happening and event B not happening,
7 × 34 = 238 or 0.7628
(c) only event B happening, and (d) either A, 8 39 312

or B, or A and B happening. Problem 7. A batch of 40 components contains 5
that are defective. If a component is drawn at
2 1 4 13 random from the batch and tested and then a second
(a) (b) (c) (d) component is drawn at random, calculate the
probability of having one defective component,
5 5 15 15 both with and without replacement

6. When testing 1000 soldered joints, 4 failed The probability of having one defective component can
be achieved in two ways. If p is the probability of draw-
during a vibration test and 5 failed due to ing a defective component and q is the probability of
drawing a satisfactory component, then the probability
having a high resistance. Determine the prob- of having one defective component is given by drawing

ability of a joint failing due to (a) vibration, (b)

high resistance, (c) vibration or high resistance

and (d) vibration and high resistance.
⎡1 1⎤
⎢⎢⎣ (a) (b)
250 200 ⎥⎥⎦
9 1
(c) (d)
1000 50 000

356 Engineering Mathematics

a satisfactory component and then a defective compo- on the first draw and the second draw and the third
nent or by drawing a defective component and then a draw is:
satisfactory one, i.e. by q × p + p × q
86 85 84 614 040
With replacement: ××=

200 199 198 7 880 400
= 0.0779

51 35 7 Problem 9. For the box of washers given in
p = = and q = = Problem 8 above, determine the probability that
40 8 40 8 there are no aluminium washers drawn, when three
washers are drawn at random from the box without
Hence, probability of having one defective component replacement
is:
The probability of not drawing an aluminium washer on
1×7+7×1
8888

i.e. 7 + 7 = 7 or 0.2188 the first draw is 1 − 40 160
64 64 32 , i.e. . There are now
200 200

199 washers in the batch of which 159 are not aluminium

Without replacement: washers. Hence, the probability of not drawing an alu-
159
p1 = 1 and q1 = 7 on the first of the two draws. The
8 8 minium washer on the second draw is . Similarly,
199
batch number is now 39 for the second draw, thus,
the probability of not drawing an aluminium washer on
p2 = 5 and q2 = 35 158
39 39
the third draw is . Hence the probability of not draw-
198

ing an aluminium washer on the first and second and

third draw is

1 35 7 5 160 × 159 × 158 = 4 019 520
p1q2 + q1p2 = 8 × 39 + 8 × 39 200 199 198 7 880 400

= 35 + 35 = 0.5101
312
Section 7 Problem 10. For the box of washers in Problem 8
70 above, find the probability that there are two brass
= or 0.2244 washers and either a steel or an aluminium washer
when three are drawn at random, without
312 replacement

Problem 8. A box contains 74 brass washer, 86 Two brass washers (A) and one steel washer (B) can be
steel washers and 40 aluminium washers. Three obtained in any of the following ways:
washers are drawn at random from the box without
replacement. Determine the probability that all
three are steel washers

Assume, for clarity of explanation, that a washer is 1st draw 2nd draw 3rd draw
A A B
drawn at random, then a second, then a third (although A B A
B A A
this assumption does not affect the results obtained).
Two brass washers and one aluminium washer (C) can
The total number of washers is 74 + 86 + 40, i.e. 200. also be obtained in any of the following ways:

The probability of randomly selecting a steel washer 1st draw 2nd draw 3rd draw
86 A A C
A C A
on the first draw is . There are now 85 steel washers C A A
200

in a batch of 199. The probability of randomly selecting
85

a steel washer on the second draw is . There are now
199

84 steel washers in a batch of 198. The probability of

randomly selecting a steel washer on the third draw is
84

. Hence the probability of selecting a steel washer
198

Probability 357

Thus there are six possible ways of achieving the com- less than 20 pins. Find the probability that
binations specified. If A represents a brass washer, B
a steel washer and C an aluminium washer, then the if 2 packets are chosen at random, one will
combinations and their probabilities are as shown:
contain less than 20 pins and the other will

contain 20 pins or more. [0.0768]

First Draw Third Probability 4. A batch of 1 kW fire elements contains 16
A Second B which are within a power tolerance and 4
A A A 74 73 86 which are not. If 3 elements are selected
B A × × = 0.0590 at random from the batch, calculate the
A B C probabilities that (a) all three are within the
A A 200 199 198 power tolerance and (b) two are within but
C A A 74 × 86 × 73 = 0.0590 one is not within the power tolerance.
200 199 198 [(a) 0.4912 (b) 0.4211]
A 86 74 73
5. An amplifier is made up of three transistors,
C × × = 0.0590
200 199 198 A, B and C. The probabilities of A, B or C
A 74 × 73 × 40 = 0.0274
200 199 198 11 1
74 × 40 × 73 = 0.0274 being defective are , and , respec-
200 199 198 20 25 50
40 × 74 × 73 = 0.0274 tively. Calculate the percentage of amplifiers
200 199 198
produced (a) which work satisfactorily and

(b) which have just one defective transistor.

The probability of having the first combination or the [(a) 89.38% (b) 10.25%]
second, or the third, and so on, is given by the sum of
the probabilities, 6. A box contains 14 40 W lamps, 28 60 W
lamps and 58 25 W lamps, all the lamps being
i.e. by 3 × 0.0590 + 3 × 0.0274, that is 0.2592 of the same shape and size. Three lamps are
drawn at random from the box, first one,
Now try the following exercise then a second, then a third. Determine the
probabilities of: (a) getting one 25 W, one
Exercise 142 Further problems on 40 W and one 60 W lamp, with replacement, Section 7
probability (b) getting one 25 W, one 40 W and one 60 W
lamp without replacement, and (c) getting
1. The probability that component A will operate either one 25 W and two 40 W or one 60 W
satisfactorily for 5 years is 0.8 and that B will and two 40 W lamps with replacement.
operate satisfactorily over that same period [(a) 0.0227 (b) 0.0234 (c) 0.0169]
of time is 0.75. Find the probabilities that in
a 5 year period: (a) both components operate 39.5 Permutations and combinations
satisfactorily, (b) only component A will
operate satisfactorily, and (c) only component Permutations
B will operate satisfactorily.
[(a) 0.6 (b) 0.2 (c) 0.15] If n different objects are available, they can be arranged
in different orders of selection. Each different ordered
2. In a particular street, 80% of the houses have arrangement is called a permutation. For example,
permutations of the three letters X, Y and Z taken
telephones. If two houses selected at random together are:

are visited, calculate the probabilities that (a) XYZ, XZY,YXZ,YZX, ZXY and ZYX

they both have a telephone and (b) one has This can be expressed as 3P3 = 6, the upper 3 denoting
the number of items from which the arrangements are
a telephone but the other does not have a made, and the lower 3 indicating the number of items
used in each arrangement.
telephone. [(a) 0.64 (b) 0.32]

3. Veroboard pins are packed in packets of 20
by a machine. In a thousand packets, 40 have

358 Engineering Mathematics

If we take the same three letters XYZ two at a time (a) 5C2 = 5! 2)! = 5!
the permutations 2!(5 − 2!3!

XY,YZ, XZ, ZX,YZ, ZY = 5×4×3×2×1 = 10
(2 × 1)(3 × 2 × 1)

can be found, and denoted by 3P2 = 6 (b) 4C2 = 4! = 4! = 6
2!(4 − 2!2!
(Note that the order of the letters matter in permutations, 2)!

i.e. YX is a different permutation from XY ). In general,

nPr = n(n − 1) (n − 2) . . . (n − r + 1) or Problem 13. A class has 24 students. 4 can
n! represent the class at an exam board. How many
nPr = (n − r)! as stated in Chapter 15 combinations are possible when choosing this group

For example, 5P4 = 5(4)(3)(2) = 120 or
5! 5!
5P4 = (5 − 4)! = 1! = (5)(4)(3)(2) = 120 Number of combinations possible,

Also, 3P3 = 6 from above; using n Pr = n! gives nCr n!
− r)! r!(n −
(n = r!)

3P3 = (3 3! = 6 . Since this must equal 6, then 24! 24!
− 3)! 0! 4!(24 − 4!20!
0! = 1 (check this with your calculator). i.e. 24C4 = = = 10 626
4)!

Combinations Problem 14. In how many ways can a team of
eleven be picked from sixteen possible players?
If selections of the three letters X, Y , Z are made with-

out regard to the order of the letters in each group, i.e.

XY is now the same as YX for example, then each group Number of ways = nCr = 16C11

is called a combination. The number of possible com- = 16! 11)! = 16! = 4368
binations is denoted by nCr, where n is the total number 11!(16 − 11!5!

of items and r is the number in each selection.

In general,

Section 7 nCr = n! r)! Now try the following exercise
r!(n −

For example,

5C4 = 5! 4)! = 5! Exercise 143 Further problems on
4!(5 − 4! permutations and
combinations
= 5 ×4×3×2× 1 = 5
4×3×2×1 1. Calculate the number of permutations there are
of: (a) 15 distinct objects taken 2 at a time,
Problem 11. Calculate the number of (b) 9 distinct objects taken 4 at a time.
permutations there are of: (a) 5 distinct objects [(a) 210 (b) 3024]
taken 2 at a time, (b) 4 distinct objects taken
2 at a time 2. Calculate the number of combinations there
are of: (a) 12 distinct objects taken 5 at a time,
(a) 5P2 = 5! = 5! = 5 × 4 × 3 × 2 = 20 (b) 6 distinct objects taken 4 at a time.
− 2)! 3! 3 × 2 [(a) 792 (b) 15]

(5 3. In how many ways can a team of six be picked

(b) 4P2 = 4! = 4! = 12 from ten possible players? [210]
− 2)! 2!
(4

4. 15 boxes can each hold one object. In how

Problem 12. Calculate the number of many ways can 10 identical objects be placed
combinations there are of: (a) 5 distinct objects taken
2 at a time, (b) 4 distinct objects taken 2 at a time in the boxes? [3003]

Revision Test 10

This Revision test covers the material in Chapters 37 to 39. The marks for each question are shown in brackets at the
end of each question.

1. A company produces five products in the follow- 4. The heights of 100 people are measured correct to
the nearest centimetre with the following results:
ing proportions:

Product A 24 Product B 16 Product C 15

Product D 11 Product E 6 150–157 cm 5 158–165 cm 18

Present these data visually by drawing (a) a ver-

tical bar chart (b) a percentage bar chart (c) a pie 166–173 cm 42 174–181 cm 27

diagram. (13) 182–189 cm 8

2. The following lists the diameters of 40 compo- Determine for the data (a) the mean height and (b)
nents produced by a machine, each measured
correct to the nearest hundredth of a centimetre: the standard deviation. (10)

1.39 1.36 1.38 1.31 1.33 1.40 1.28 5. Determine the probabilities of:
1.40 1.24 1.28 1.42 1.34 1.43 1.35
1.36 1.36 1.35 1.45 1.29 1.39 1.38 (a) drawing a white ball from a bag containing
1.38 1.35 1.42 1.30 1.26 1.37 1.33
1.37 1.34 1.34 1.32 1.33 1.30 1.38 6 black and 14 white balls
1.41 1.35 1.38 1.27 1.37
(b) winning a prize in a raffle by buying 6 tickets

when a total of 480 tickets are sold

(c) selecting at random a female from a group

of 12 boys and 28 girls

(d) winning a prize in a raffle by buying 8 tickets

when there are 5 prizes and a total of 800

tickets are sold. (8)

(a) Using 8 classes form a frequency distribution and 6. In a box containing 120 similar transistors 70 are Section 7
a cumulative frequency distribution. satisfactory, 37 give too high a gain under normal
operating conditions and the remainder give too
(b) For the above data draw a histogram, a frequency low a gain.

polygon and an ogive. (21) Calculate the probability that when drawing two
transistors in turn, at random, with replacement,
3. Determine for the 10 measurements of lengths of having (a) two satisfactory, (b) none with low
shown below: gain, (c) one with high gain and one satisfactory,
(d) one with low gain and none satisfactory.
(a) the arithmetic mean, (b) the median, (c) the
mode, and (d) the standard deviation. Determine the probabilities in (a), (b) and

28 m, 20 m, 32 m, 44 m, 28 m, 30 m, 30 m, 26 m, (c) above if the transistors are drawn without

28 m and 34 m (9) replacement. (14)

Chapter 40

The binomial and Poisson
distribution

40.1 The binomial distribution a family of 4 children, assuming equal probability
of male and female birth
The binomial distribution deals with two numbers only,
these being the probability that an event will happen, The probability of a girl being born, p, is 0.5 and the
p, and the probability that an event will not happen, q. probability of a girl not being born (male birth), q, is
Thus, when a coin is tossed, if p is the probability of the also 0.5. The number in the family, n, is 4. From above,
coin landing with a head upwards, q is the probability of the probabilities of 0, 1, 2, 3, 4 girls in a family of 4
the coin landing with a tail upwards. p + q must always are given by the successive terms of the expansion of
be equal to unity. A binomial distribution can be used (q + p)4 taken from left to right.
for finding, say, the probability of getting three heads in
seven tosses of the coin, or in industry for determining From the binomial expansion:
defect rates as a result of sampling. One way of defining (q + p)4 = q4 + 4q3p + 6q2p2 + 4qp3 + p4
a binomial distribution is as follows:
Hence the probability of no girls is q4,
‘if p is the probability that an event will happen and i.e. 0.54 = 0.0625
q is the probability that the event will not happen, the probability of 1 girl is 4q3p,
then the probabilities that the event will happen 0, i.e. 4 × 0.53 × 0.5 = 0.2500
1, 2, 3, . . ., n times in n trials are given by the the probability of 2 girls is 6q2p2,
successive terms of the expansion of (q + p)n taken i.e. 6 × 0.52 × 0.52 = 0.3750
from left to right’. the probability of 3 girls is 4qp3,
i.e. 4 × 0.5 × 0.53 = 0.2500
The binomial expansion of (q + p)n is: the probability of 4 girls is p4,
i.e. 0.54_=__0_._0_6_2_5
qn + nqn−1p + n(n − 1) qn−2p2
2! Total probability, (q + p)4_=__1_._0_0_0_0

+ n(n − 1)(n − 2) qn−3p3 + · · · (a) The probability of having at least one girl is the
3! sum of the probabilities of having 1, 2, 3 and 4
girls, i.e.
from Chapter 16
This concept of a binomial distribution is used in 0.2500 + 0.3750 + 0.2500 + 0.0625 = 0.9375

Problems 1 and 2.

Problem 1. Determine the probabilities of having
(a) at least 1 girl and (b) at least 1 girl and 1 boy in

The binomial and Poisson distribution 361

(Alternatively, the probability of having at least 1 Industrial inspection Section 7
girl is: 1 – (the probability of having no girls), i.e.
1 – 0.0625, giving 0.9375, as obtained previously). In industrial inspection, p is often taken as the prob-
ability that a component is defective and q is the
(b) The probability of having at least 1 girl and 1 boy probability that the component is satisfactory. In this
is given by the sum of the probabilities of having: case, a binomial distribution may be defined as:
1 girl and 3 boys, 2 girls and 2 boys and 3 girls and
2 boys, i.e. ‘the probabilities that 0, 1, 2, 3, …, n components
are defective in a sample of n components, drawn
0.2500 + 0.3750 + 0.2500 = 0.8750 at random from a large batch of components, are
given by the successive terms of the expansion of
(Alternatively, this is also the probability of having (q + p)n, taken from left to right’.
1 − (probability of having no girls + probability of
having no boys), i.e. 1 − 2 × 0.0625 = 0.8750, as This definition is used in Problems 3 and 4.
obtained previously).
Problem 3. A machine is producing a large
Problem 2. A dice is rolled 9 times. Find the number of bolts automatically. In a box of these
probabilities of having a 4 upwards (a) 3 times and bolts, 95% are within the allowable tolerance values
(b) less than 4 times with respect to diameter, the remainder being
outside of the diameter tolerance values. Seven
Let p be the probability of having a 4 upwards. Then bolts are drawn at random from the box. Determine
p = 1/6, since dice have six sides. the probabilities that (a) two and (b) more than two
of the seven bolts are outside of the diameter
Let q be the probability of not having a 4 upwards. tolerance values
Then q = 5/6. The probabilities of having a 4 upwards
0, 1, 2. . . n times are given by the successive terms of Let p be the probability that a bolt is outside of the
the expansion of (q + p)n, taken from left to right. From allowable tolerance values, i.e. is defective, and let q be
the binomial expansion: the probability that a bolt is within the tolerance values,
i.e. is satisfactory. Then p = 5%, i.e. 0.05 per unit and
(q + q)9 = q9 + 9q8p + 36q7p2 + 84q6p3 + . . . q = 95%, i.e. 0.95 per unit. The sample number is 7.

The probability of having a 4 upwards no times is The probabilities of drawing 0, 1, 2, . . . , n defective
bolts are given by the successive terms of the expansion
q9 = (5/6)9 = 0.1938 of (q + p)n, taken from left to right. In this problem

The probability of having a 4 upwards once is (q + p)n = (0.95 + 0.05)7
= 0.957 + 7 × 0.956 × 0.05
9q8p = 9(5/6)8(1/6) = 0.3489 + 21 × 0.955 × 0.052 + . . .

The probability of having a 4 upwards twice is Thus the probability of no defective bolts is:

36q7p2 = 36(5/6)7(1/6)2 = 0.2791 0.957 = 0.6983

The probability of having a 4 upwards 3 times is The probability of 1 defective bolt is:

84q6p3 = 84(5/6)6(1/6)3 = 0.1302 7 × 0.956 × 0.05 = 0.2573

(a) The probability of having a 4 upwards 3 times is The probability of 2 defective bolts is:
0.1302
21 × 0.955 × 0.052 = 0.0406, and so on.
(b) The probability of having a 4 upwards less than 4
times is the sum of the probabilities of having a 4 (a) The probability that two bolts are outside of the
upwards 0, 1, 2, and 3 times, i.e. diameter tolerance values is 0.0406

0.1938 + 0.3489 + 0.2791 + 0.1302 = 0.9520 (b) To determine the probability that more than two
bolts are defective, the sum of the probabilities

362 Engineering Mathematics

of 3 bolts, 4 bolts, 5 bolts, 6 bolts and 7 boltsSection 7 Let p be the probability of a student successfully com-
being defective can be determined. An easier way pleting a course of study in three years and q be the
to find this sum is to find 1 − (sum of 0 bolts, 1 bolt Probability of successfully completing courseprobability of not doing so. Then p = 0.45 and q = 0.55.
and 2 bolts being defective), since the sum of all The number of students, n, is 10.
the terms is unity. Thus, the probability of there
being more than two bolts outside of the tolerance The probabilities of 0, 1, 2, . . ., 10 students success-
values is: fully completing the course are given by the successive
terms of the expansion of (q + p)10, taken from left
1 − (0.6983 + 0.2573 + 0.0406), i.e. 0.0038 to right.

Problem 4. A package contains 50 similar (q + p)10 = q10 + 10q9p + 45q8p2 + 120q7p3
components and inspection shows that four have + 210q6p4 + 252q5p5 + 210q4p6
been damaged during transit. If six components are + 120q3p7 + 45q2p8 + 10qp9 + p10
drawn at random from the contents of the package
determine the probabilities that in this sample Substituting q = 0.55 and p = 0.45 in this expansion
(a) one and (b) less than three are damaged gives the values of the success terms as:

The probability of a component being damaged, p, is 0.0025, 0.0207, 0.0763, 0.1665, 0.2384, 0.2340,
4 in 50, i.e. 0.08 per unit. Thus, the probability of a 0.1596, 0.0746, 0.0229, 0.0042 and 0.0003. The
component not being damaged, q, is 1 − 0.08, i.e. 0.92 histogram depicting these probabilities is shown in
The probability of there being 0, 1, 2, . . ., 6 damaged Fig. 40.1.
components is given by the successive terms of (q + p)6,
taken from left to right. 0.24

(q + p)6 = q6 + 6q5p + 15q4p2 + 20q3p3 + · · · 0.22

(a) The probability of one damaged component is 0.20

6q5p = 6 × 0.925 × 0.08 = 0.3164 0.18

(b) The probability of less than three damaged com- 0.16
ponents is given by the sum of the probabilities of
0, 1 and 2 damaged components. 0.14

q6 + 6q5p + 15q4p2 0.12
= 0.926 + 6 × 0.925 × 0.08
+ 15 × 0.924 × 0.082 0.10
= 0.6064 + 0.3164 + 0.0688 = 0.9916
0.08
Histogram of probabilities
0.06
The terms of a binomial distribution may be repre-
sented pictorially by drawing a histogram, as shown 0.04
in Problem 5.
0.02
Problem 5. The probability of a student
successfully completing a course of study in three 0
years is 0.45. Draw a histogram showing the 0 1 2 3 4 5 6 7 8 9 10
probabilities of 0, 1, 2, . . ., 10 students successfully Number of students
completing the course in three years
Figure 40.1

Now try the following exercise The binomial and Poisson distribution 363

Exercise 144 Further problems on the 7. An automatic machine produces, on average,
binomial distribution 10% of its components outside of the tolerance
required. In a sample of 10 components from
1. Concrete blocks are tested and it is found this machine, determine the probability of hav-
that, on average, 7% fail to meet the required ing three components outside of the tolerance
specification. For a batch of 9 blocks, deter- required by assuming a binomial distribution.
mine the probabilities that (a) three blocks [0.0574]
and (b) less than four blocks will fail to meet
the specification. 40.2 The Poisson distribution
[(a) 0.0186 (b) 0.9976]

2. If the failure rate of the blocks in Problem 1 When the number of trials, n, in a binomial distribution
rises to 15%, find the probabilities that (a) no becomes large (usually taken as larger than 10), the cal-
blocks and (b) more than two blocks will fail culations associated with determining the values of the
to meet the specification in a batch of 9 blocks. terms become laborious. If n is large and p is small, and
[(a) 0.2316 (b) 0.1408] the product np is less than 5, a very good approximation
to a binomial distribution is given by the corresponding
3. The average number of employees absent Poisson distribution, in which calculations are usually
from a firm each day is 4%. An office within simpler.
the firm has seven employees. Determine the
probabilities that (a) no employee and (b) three The Poisson approximation to a binomial distribution
employees will be absent on a particular day. may be defined as follows:
[(a) 0.7514 (b) 0.0019]
‘the probabilities that an event will happen 0, 1, 2,
4. A manufacturer estimates that 3% of his 3, . . . , n times in n trials are given by the successive
output of a small item is defective. Find the terms of the expression
probabilities that in a sample of 10 items (a)
less than two and (b) more than two items will e−λ λ2 λ3 Section 7
be defective. 1+λ+ + +···
[(a) 0.9655 (b) 0.0028] 2! 3!

5. Five coins are tossed simultaneously. Deter- taken from left to right’

mine the probabilities of having 0, 1, 2, 3, 4 The symbol λ is the expectation of an event happening
and is equal to np.
and 5 heads upwards, and draw a histogram

depicting the results.
⎡⎤
Vertical adjacent rectangles, Problem 6. If 3% of the gearwheels produced
⎢⎢⎣ ⎥⎦⎥ by a company are defective, determine the
whose heights are proportional to probabilities that in a sample of 80 gearwheels
0.0313, 0.1563, 0.3125, 0.3125, (a) two and (b) more than two will be defective.

0.1563 and 0.0313

6. If the probability of rain falling during a The sample number, n, is large, the probability of a
defective gearwheel, p, is small and the product np is
particular period is 2/5, find the probabilities 80 × 0.03, i.e. 2.4, which is less than 5. Hence a Poisson
approximation to a binomial distribution may be used.
of having 0, 1, 2, 3, 4, 5, 6 and 7 wet days in The expectation of a defective gearwheel, λ = np = 2.4

a week. Show these results on a histogram. The probabilities of 0, 1, 2, . . . defective gearwheels
⎡⎤ are given by the successive terms of the expression
Vertical adjacent rectangles,
⎣⎢⎢⎢⎢ ⎥⎥⎥⎥⎦
whose heights are proportional
to 0.0280, 0.1306, 0.2613,
0.2903, 0.1935, 0.0774,

0.0172 and 0.0016 e−λ 1+ λ + λ2 + λ3 +···
2! 3!

364 Engineering Mathematics

taken from left to right, i.e. by Since the average occurrence of a breakdown is known
e−λ, λe−λ, λ2e−λ , . . . Thus: but the number of times when a machine did not break
2! down is unknown, a Poisson distribution must be used.

probability of no defective gearwheels is The expectation of a breakdown for 35 machines is
e−λ = e−2.4 = 0.0907 35 × 0.06, i.e. 2.1 breakdowns per week. The probabili-
ties of a breakdown occurring 0, 1, 2, . . . times are given
probability of 1 defective gearwheel is by the successive terms of the expression
λe−λ = 2.4e−2.4 = 0.2177
e−λ 1 + λ + λ2 + λ3 + · · · ,
probability of 2 defective gearwheels is 2! 3!
λ2e−λ 2.42e−2.4
2! = 2 × 1 = 0.2613 taken from left to right. Hence:

(a) The probability of having 2 defective gearwheels probability of no breakdowns
is 0.2613
e−λ = e−2.1 = 0.1225
(b) The probability of having more than 2 defective
gearwheels is 1 − (the sum of the probabilities of probability of 1 breakdown is
having 0, 1, and 2 defective gearwheels), i.e.
λe−λ = 2.1e−2.1 = 0.2572
1 − (0.0907 + 0.2177 + 0.2613),
probability of 2 breakdowns is
that is, 0.4303
λ2e−λ 2.12e−2.1
Section 7 The principal use of a Poisson distribution is to deter- = = 0.2700
mine the theoretical probabilities when p, the prob-
ability of an event happening, is known, but q, the 2! 2 × 1
probability of the event not happening is unknown. For
example, the average number of goals scored per match (a) The probability of 1 breakdown per week is 0.2572
by a football team can be calculated, but it is not possi-
ble to quantify the number of goals that were not scored. (b) The probability of less than 3 breakdowns per
In this type of problem, a Poisson distribution may be week is the sum of the probabilities of 0, 1 and
defined as follows: 2 breakdowns per week,
i.e. 0.1225 + 0.2572 + 0.2700 = 0.6497

‘the probabilities of an event occurring 0, 1, 2, Histogram of probabilities
3…times are given by the successive terms of the
expression The terms of a Poisson distribution may be repre-
sented pictorially by drawing a histogram, as shown
e−λ λ2 λ3 , in Problem 8.
1+λ+ + +···
2! 3! Problem 8. The probability of a person having an
accident in a certain period of time is 0.0003. For
taken from left to right’ a population of 7500 people, draw a histogram
showing the probabilities of 0, 1, 2, 3, 4, 5 and 6
people having an accident in this period.

The symbol λ is the value of the average occurrence of The probabilities of 0, 1, 2, . . . people having an
the event. accident are given by the terms of the expression

Problem 7. A production department has 35 e−λ λ2 λ3 ,
similar milling machines. The number of 1+λ+ + +···
breakdowns on each machine averages 0.06 per 2! 3!
week. Determine the probabilities of having
(a) one, and (b) less than three machines breaking taken from left to right.
down in any week The average occurrence of the event, λ, is

7500 × 0.0003, i.e. 2.25

The binomial and Poisson distribution 365

The probability of no people having an accident is 2. The probability that an employee will go to
hospital in a certain period of time is 0.0015.
e−λ = e−2.25 = 0.1054 Use a Poisson distribution to determine the
probability of more than two employees going
The probability of 1 person having an accident is to hospital during this period of time if there
are 2000 employees on the payroll.
λe−λ = 2.25e−2.25 = 0.2371 [0.5768]

The probability of 2 people having an accident is 3. When packaging a product, a manufacturer

λ2e−λ 2.252e−2.25 finds that one packet in twenty is underweight.
= = 0.2668
Determine the probabilities that in a box of 72
2! 2!
and so on, giving probabilities of 0.2001, 0.1126, packets (a) two and (b) less than four will be
0.0506 and 0.0190 for 3, 4, 5 and 6 respectively hav-
ing an accident. The histogram for these probabilities is underweight. [(a) 0.1771 (b) 0.5153]
shown in Fig. 40.2.
4. A manufacturer estimates that 0.25% of his
0.28 output of a component are defective. The com-
ponents are marketed in packets of 200. Deter-
0.24 mine the probability of a packet containing less
than three defective components.
[0.9856]
Probability of having an accident

Section 7
0.20 5. The demand for a particular tool from a store

is, on average, five times a day and the demand

0.16 follows a Poisson distribution. How many of

these tools should be kept in the stores so that

0.12 the probability of there being one available

when required is greater than 10%?
⎡⎤
The probabilities of the demand
0.08 ⎢⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢ ⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

0.04 for 0, 1, 2, . . . tools are
0.0067, 0.0337, 0.0842, 0.1404,
0 1 2 34 5 6 0.1755, 0.1755, 0.1462, 0.1044,
0 Number of people 0.0653, . . . This shows that the
probability of wanting a tool
Figure 40.2 8 times a day is 0.0653, i.e.
less than 10%. Hence 7 should

be kept in the store

Now try the following exercise 6. Failure of a group of particular machine tools

follows a Poisson distribution with a mean

value of 0.7. Determine the probabilities of 0,

Exercise 145 Further problems on the 1, 2, 3, 4 and 5 failures in a week and present
Poisson distribution
these results on a histogram.
⎡⎤
1. In problem 7 of Exercise 144, page 363, Vertical adjacent rectangles
⎢⎢⎣ ⎥⎥⎦
determine the probability of having three com- having heights proportional
to 0.4966, 0.3476, 0.1217,
ponents outside of the required tolerance using
0.0284, 0.0050 and 0.0007
the Poisson distribution. [0.0613]

Chapter 41

The normal distribution

41.1 Introduction to the normal and (d) by having different areas between the curve
distribution and the horizontal axis.

When data is obtained, it can frequently be considered A normal distribution curve is standardised as
to be a sample (i.e. a few members) drawn at random follows:
from a large population (i.e. a set having many mem-
bers). If the sample number is large, it is theoretically (a) The mean value of the unstandardised curve is made
possible to choose class intervals which are very small, the origin, thus making the mean value, x¯, zero.
but which still have a number of members falling within
each class. A frequency polygon of this data then has (b) The horizontal axis is scaled in standard deviations.
a large number of small line segments and approxi- This is done by letting z = x − x¯ , where z is called
mates to a continuous curve. Such a curve is called a σ
frequency or a distribution curve. the normal standard variate, x is the value of the
variable, x¯ is the mean value of the distribution and
An extremely important symmetrical distribution σ is the standard deviation of the distribution.
curve is called the normal curve and is as shown in
Fig. 41.1. This curve can be described by a mathemati- (c) The area between the normal curve and the hori-
cal equation and is the basis of much of the work done zontal axis is made equal to unity.
in more advanced statistics. Many natural occurrences
such as the heights or weights of a group of people, the When a normal distribution curve has been standard-
sizes of components produced by a particular machine ised, the normal curve is called a standardised normal
and the life length of certain components approximate curve or a normal probability curve, and any normally
to a normal distribution. distributed data may be represented by the same normal
probability curve.
Frequency
The area under part of a normal probability curve is
Variable directly proportional to probability and the value of the
shaded area shown in Fig. 41.2 can be determined by
evaluating:

√1 e(z2/2)dz, where z = x − x¯
2π σ

Probability
density

Figure 41.1

Normal distribution curves can differ from one another z1 0 z2 z-value
in the following four ways: Standard deviations

(a) by having different mean values Figure 41.2
(b) by having different values of standard

deviations
(c) the variables having different values and

different units

The normal distribution 367

To save repeatedly determining the values of this func- Assuming the heights are normally distributed,
tion, tables of partial areas under the standardised nor- determine the number of people likely to have
mal curve are available in many mathematical formulae heights between 150 cm and 195 cm
books, and such a table is shown in Table 41.1.
The mean value, x¯, is 170 cm and corresponds to a nor-
Problem 1. The mean height of 500 people is mal standard variate value, z, of zero on the standardised
170 cm and the standard deviation is 9 cm. normal curve. A height of 150 cm has a z-value given

Table 41.1 Partial areas under the standardised normal curve

0z Section 7

z = x − x¯ 0 1 2 3 4 5 6 7 8 9
σ

0.0 0.0000 0.0040 0.0080 0.0120 0.0159 0.0199 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0678 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1388 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2086 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2760 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3451 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4430 0.4441

(Continued )

368 Engineering Mathematics

Section 7 Table 41.1 (Continued )
z = x − x¯ 0 1 2 3 4 5 6 7 8 9
σ
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4762 0.4767
2.0 0.4772 0.4778 0.4783 0.4785 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964
2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974
2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4980 0.4980 0.4981
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
3.1 0.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.4993
3.2 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.4995
3.3 0.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.4997
3.4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4998
3.5 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998
3.6 0.4998 0.4998 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999
3.7 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999
3.8 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999
3.9 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 05000 0.5000 0.5000 0.5000

by z = x − x¯ standard deviations, i.e. 150 − 170 or The negative z-value shows that it lies to the left of the
σ9
z = 0 ordinate.
−2.22 standard deviations. Using a table of partial areas
beneath the standardised normal curve (see Table 41.1), This area is shown shaded in Fig. 41.3(a). Similarly,
a z-value of −2.22 corresponds to an area of 0.4868
between the mean value and the ordinate z = −2.22. 195 cm has a z-value of 195 − 170 that is 2.78 standard

9
deviations. From Table 41.1, this value of z corresponds

The normal distribution 369

Ϫ2.22 0 z-value Ϫ0.56 0 z-value
(a) (a)

0 2.78 z-value Ϫ0.56 0 z-value
(b) (b)

Figure 41.4

Ϫ2.22 0 2.78 z-value follows that the total area to the left of the z = 0 ordinate
(c) is 0.5000. Thus the area to the left of the z = −0.56 ordi-
nate (‘left’ means ‘less than’, ‘right’ means ‘more than’)
Figure 41.3 is 0.5000 − 0.2123, i.e. 0.2877 of the total area, which
is shown shaded in Fig. 41.4(b). The area is directly pro-
to an area of 0.4973, the positive value of z showing portional to probability and since the total area beneath
that it lies to the right of the z = 0 ordinate. This area the standardised normal curve is unity, the probability of
a person’s height being less than 165 cm is 0.2877. For
is show shaded in Fig. 41.3(b). The total area shaded a group of 500 people, 500 × 0.2877, i.e. 144 people
in Fig. 41.3(a) and (b) is shown in Fig. 41.3(c) and is are likely to have heights of less than 165 cm.
0.4868 + 0.4973, i.e. 0.9841 of the total area beneath
the curve. Problem 3. For the group of people given in Section 7
Problem 1 find how many people are likely to have
However, the area is directly proportional to probabil- heights of more than 194 cm
ity. Thus, the probability that a person will have a height
of between 150 and 195 cm is 0.9841. For a group of 194 cm correspond to a z-value of 194 − 170 that is,
500 people, 500 × 0.9841, i.e. 492 people are likely to
have heights in this range. The value of 500 × 0.9841 9
is 492.05, but since answers based on a normal proba- 2.67 standard deviations. From Table 41.1, the area
bility distribution can only be approximate, results are
usually given correct to the nearest whole number. between z = 0, z = 2.67 and the standardised normal

curve is 0.4962, shown shaded in Fig. 41.5(a). Since

the standardised normal curve is symmetrical, the total

area to the right of the z = 0 ordinate is 0.5000, hence the

shaded area shown in Fig. 41.5(b) is 0.5000 − 0.4962,

i.e. 0.0038. This area represents the probability of

Problem 2. For the group of people given in a person having a height of more than 194 cm, and
Problem 1, find the number of people likely to have
heights of less than 165 cm for 500 people, the number of people likely to have

a height of more than 194 cm is 0.0038 × 500, i.e.

2 people.

A height of 165 cm corresponds to 165 − 170 , i.e.

9 Problem 4. A batch of 1500 lemonade bottles
−0.56 standard deviations. The area between z = 0 and have an average contents of 753 ml and the standard
deviation of the contents is 1.8 ml. If the volumes of
z = −0.56 (from Table 41.1) is 0.2123, shown shaded in the content are normally distributed, find the

Fig. 41.4(a). The total area under the standardised nor-

mal curve is unity and since the curve is symmetrical, it

370 Engineering Mathematics

0 2.67 z-value 0.5000 − 0.4868 (see Problem 3), i.e. 0.0132 Thus,
(a) for 1500 bottles, it is likely that 1500 × 0.0132,
i.e. 20 bottles will have contents of more
0 2.67 z-value than 750 ml.
(b)
(d) The z-value corresponding to 750 ml is −1.67 (see
Figure 41.5 part (a)), and the z-value corresponding to 751 ml
is −1.11 (see part (b)). The areas corresponding to
these z-values area 0.4525 and 0.3665 respectively,
and both these areas lie on the left of the z = 0 ordi-
nate. The area between z = −1.67 and z = −1.11
is 0.4525 − 0.3665, i.e. 0.0860 and this is the
probability of a bottle having contents between
750 and 751 ml. For 1500 bottles, it is likely that
1500 × 0.0860, i.e. 129 bottles will be in this
range.

Now try the following exercise

number of bottles likely to contain: (a) less than Exercise 146 Further problems on the
750 ml, (b) between 751 and 754 ml, (c) more than introduction to the normal
757 ml, and (d) between 750 and 751 ml distribution

1. A component is classed as defective if it has a

diameter of less than 69 mm. In a batch of 350

(a) The z-value corresponding to 750 ml is given components, the mean diameter is 75 mm and
by x − x¯ i.e. 750 − 753 = −1.67 standard devi-
σ 1.8 the standard deviation is 2.8 mm. Assuming
ations. From Table 41.1, the area between z = 0
and z = −1.67 is 0.4525. Thus the area to the the diameters are normally distributed, deter-
left of the z = −1.67 ordinate is 0.5000 − 0.4525
Section 7 (see Problem 2), i.e. 0.0475. This is the prob- mine how many are likely to be classed as

ability of a bottle containing less than 750 ml. defective [6]

Thus, for a batch of 1500 bottles, it is likely that 2. The masses of 800 people are normally dis-
1500 × 0.0475, i.e. 71 bottles will contain less
than 750 ml. tributed, having a mean value of 64.7 kg, and

a standard deviation of 5.4 kg. Find how many

people are likely to have masses of less than

54.4 kg. [22]

(b) The z-value corresponding to 751 and 754 ml 3. 500 tins of paint have a mean content of
are 751 − 753 and 754 − 753 i.e. −1.11 and 1010 ml and the standard deviation of the con-
1.8 1.8 tents is 8.7 ml. Assuming the volumes of the
0.56 respectively. From Table 41.1, the areas contents are normally distributed, calculate
the number of tins likely to have contents
corresponding to these values are 0.3665 and whose volumes are less than (a) 1025 ml (b)
1000 ml and (c) 995 ml.
0.2123 respectively. Thus the probability of a [(a) 479 (b) 63 (c) 21]

bottle containing between 751 and 754 ml is

0.3665 + 0.2123 (see Problem 1), i.e. 0.5788.

For 1500 bottles, it is likely that 1500 × 0.5788,

i.e. 868 bottles will contain between 751

and 754 ml. 4. For the 350 components in Problem 1, if those

(c) The z-value corresponding to 757 ml is 757 − 753 , having a diameter of more than 81.5 mm are

1.8 rejected, find, correct to the nearest compo-

i.e. 2.22 standard deviations. From Table 41.1, the nent, the number likely to be rejected due to

area corresponding to a z-value of 2.22 is 0.4868. being oversized. [4]

The area to the right of the z = 2.22 ordinate is

The normal distribution 371

5. For the 800 people in Problem 2, determine is shown in Problems 5 and 6. The mean value and
how many are likely to have masses of more standard deviation of normally distributed data may be
than (a) 70 kg, and (b) 62 kg. determined using normal probability paper. For nor-
[(a) 131 (b) 553] mally distributed data, the area beneath the standardised
normal curve and a z-value of unity (i.e. one standard
6. The mean diameter of holes produced by a deviation) may be obtained from Table 41.1. For one
drilling machine bit is 4.05 mm and the stan- standard deviation, this area is 0.3413, i.e. 34.13%. An
dard deviation of the diameters is 0.0028 mm. area of ±1 standard deviation is symmetrically placed
For twenty holes drilled using this machine, on either side of the z = 0 value, i.e. is symmetrically
determine, correct to the nearest whole num- placed on either side of the 50 per cent cumulative
ber, how many are likely to have diame- frequency value. Thus an area corresponding to ±1
ters of between (a) 4.048 and 4.0553 mm, standard deviation extends from percentage cumulative
and (b) 4.052 and 4.056 mm, assuming the frequency values of (50 + 34.13)% to (50 − 34.13)%,
diameters are normally distributed. i.e. from 84.13% to 15.87%. For most purposes, these
[(a) 15 (b) 4] values area taken as 84% and 16%. Thus, when using
normal probability paper, the standard deviation of the
7. The intelligence quotients of 400 children distribution is given by:
have a mean value of 100 and a standard
deviation of 14. Assuming that I.Q.’s are nor- (variable value for 84% cumulative frequency) −
mally distributed, determine the number of
children likely to have I.Q.’s of between (a) 80 (variable value for 16% cumulative frequency)
and 90, (b) 90 and 110, and (c) 110 and 130.
[(a) 65 (b) 209 (c) 89] 2

8. The mean mass of active material in tablets Problem 5. Use normal probability paper to
produced by a manufacturer is 5.00 g and the determine whether the data given below, which
standard deviation of the masses is 0.036 g. In refers to the masses of 50 copper ingots, is
a bottle containing 100 tablets, find how many approximately normally distributed. If the data is
tablets are likely to have masses of (a) between normally distributed, determine the mean and
4.88 and 4.92 g, (b) between 4.92 and 5.04 g, standard deviation of the data from the graph drawn
and (c) more than 5.04 g.
[(a) 1 (b) 85 (c) 13] Class mid-point Section 7

41.2 Testing for a normal distribution value (kg) 29.5 30.5 31.5 32.5 33.5

It should never be assumed that because data is con- Frequency 24689
tinuous it automatically follows that it is normally
distributed. One way of checking that data is normally Class mid-point
distributed is by using normal probability paper, often
just called probability paper. This is special graph value (kg) 34.5 35.5 36.5 37.5 38.5
paper which has linear markings on one axis and per-
centage probability values from 0.01 to 99.99 on the Frequency 86421
other axis (see Figs. 41.6 and 41.7). The divisions on the
probability axis are such that a straight line graph results To test the normality of a distribution, the upper class
for normally distributed data when percentage cumu-
lative frequency values are plotted against upper class boundary/percentage cumulative frequency values are
boundary values. If the points do not lie in a reasonably
straight line, then the data is not normally distributed. plotted on normal probability paper. The upper class
The method used to test the normality of a distribution
boundary values are: 30, 31, 32, . . . , 38, 39. The cor-

responding cumulative frequency values (for ‘less than’
the upper class boundary values) are: 2, (4 + 2) = 6,
(6 + 4 + 2) = 12, 20, 29, 37, 43, 47, 49 and 50. The
corresponding percentage cumulative frequency values
are 2 × 100 = 4, 6 × 100 = 12, 24, 40, 58, 74, 86,

50 50
94, 98 and 100%.

The co-ordinates of upper class boundary/percentage

cumulative frequency values are plotted as shown in

372 Engineering Mathematics

99.99 standard deviation, σ = [f (x − x¯)2]
where f is the
99.9
99.8 f

99 frequency of a class and x is the class mid-point value.
98
Using these formulae gives a mean value of the distri-
95
bution of 33.6 (as obtained graphically) and a standard
90
Q deviation of 2.12, showing that the graphical method of

80 determining the mean and standard deviation give quite

70Percentage cumulative frequency realistic results.
60
50 P Problem 6. Use normal probability paper to
40 determine whether the data given below is normally
30 distributed. Use the graph and assume a normal
20 R distribution whether this is so or not, to find
approximate values of the mean and standard
10 deviation of the distribution.

5 Class mid-point

2 Values 5 15 25 35 45
1
0.5 Frequency 12369
0.2
0.1 Class mid-point
0.05
Values 55 65 75 85 95
0.01
30 32 34 36 38 40 42 Frequency 62211
Upper class boundary
Section 7 To test the normality of a distribution, the upper class
Figure 41.6 boundary/percentage cumulative frequency values are
plotted on normal probability paper. The upper class
Fig. 41.6. When plotting these values, it will always boundary values are: 10, 20, 30, . . . , 90 and 100.
be found that the co-ordinate for the 100% cumulative The corresponding cumulative frequency values are 1,
frequency value cannot be plotted, since the maximum 1 + 2 = 3, 1 + 2 + 3 = 6, 12, 21, 27, 29, 31, 32 and
value on the probability scale is 99.99. Since the points 33. The percentage cumulative frequency values are
plotted in Fig. 41.6 lie very nearly in a straight line, 1 × 100 = 3, 3 × 100 = 9, 18, 36, 64, 82, 88, 94,
the data is approximately normally distributed. 33 33
97 and 100.
The mean value and standard deviation can be deter-
mined from Fig. 41.6. Since a normal curve is sym- The co-ordinates of upper class boundary values/
metrical, the mean value is the value of the variable percentage cumulative frequency values are plotted as
corresponding to a 50% cumulative frequency value, shown in Fig. 41.7. Although six of the points lie approx-
shown as point P on the graph. This shows that the imately in a straight line, three points corresponding to
mean value is 33.6 kg. The standard deviation is deter- upper class boundary values of 50, 60 and 70 are not
mined using the 84% and 16% cumulative frequency close to the line and indicate that the distribution is
values, shown as Q and R in Fig. 41.6. The variable val- not normally distributed. However, if a normal dis-
ues for Q and R are 35.7 and 31.4 respectively; thus two tribution is assumed, the mean value corresponds to
standard deviations correspond to 35.7 − 31.4, i.e. 4.3, the variable value at a cumulative frequency of 50%
showing that the standard deviation of the distribution and, from Fig. 41.7, point A is 48. The value of the
standard deviation of the distribution can be obtained
4.3 from the variable values corresponding to the 84% and
is approximately i.e. 2.15 standard deviations. 16% cumulative frequency values, shown as B and C
in Fig. 41.7 and give: 2σ = 69 − 28, i.e. the standard
2
The mean value and standard deviation of the

fx
distribution can be calculated using mean, x¯ = and

f

The normal distribution 373

99.99 Class mid-point
99.9

value 27.2 27.4 27.6

99 Frequency 36 25 12
98
Use normal probability paper to show that this
95 data approximates to a normal distribution and
90 hence determine the approximate values of the
mean and standard deviation of the distribu-
80 tion. Use the formula for mean and standard
70 deviation to verify the results obtained.
60
50 ⎡⎤
40 ⎣⎢Graphically, x¯ = 27.1, σ = 0.3; by ⎦⎥
30
20 calculation, x¯ = 27.079, σ = 0.3001

10 2. A frequency distribution of the class mid-point
5 values of the breaking loads for 275 similar
fibres is as shown below:
2 Percentage cumulative frequency B
1 A
0.5 Section 7C
0.2 Load (kN) 17 19 21 23
0.1 10 20 30 40 50 60 70 80 90 100 110 Frequency 9 23 55 78
0.05 Upper class boundary
Load (kN) 25 27 29 31
0.01 Frequency 64 28 14 4

Figure 41.7

deviation σ = 20.5. The calculated values of the mean Use normal probability paper to show that this
and standard deviation of the distribution are 45.9 and
19.4 respectively, showing that errors are introduced if distribution is approximately normally dis-
the graphical method of determining these values is used
for data that is not normally distributed. tributed and determine the mean and standard

deviation of the distribution (a) from the graph

and (b) by calculation.

⎡⎤

Now try the following exercise ⎣⎢ (a) x¯ = 23.5 kN, σ = 2.9 kN ⎦⎥

(b) x¯ = 23.364 kN, σ = 2.917 kN

Exercise 147 Further problems on testing
for a normal distribution

1. A frequency distribution of 150 measurements
is as shown:

Class mid-point

value 26.4 26.6 26.8 27.0

Frequency 5 12 24 36

Revision Test 11

This Revision test covers the material contained in Chapters 40 and 41. The marks for each question are shown in
brackets at the end of each question.

1. A machine produces 15% defective components. In a batch of 500 components, determine the
In a sample of 5, drawn at random, calculate, using number of components likely to:
the binomial distribution, the probability that:
(a) have a length of less than 19.95 mm

(a) there will be 4 defective items (b) be between 19.95 mm and 20.15 mm

(b) there will be not more than 3 defective items (c) be longer than 20.54 mm. (12)

(c) all the items will be non-defective (14)

2. 2% of the light bulbs produced by a company 4. In a factory, cans are packed with an average of
are defective. Determine, using the Poisson dis-
tribution, the probability that in a sample of 80 1.0 kg of a compound and the masses are normally
bulbs:
distributed about the average value. The standard

deviation of a sample of the contents of the cans

is 12 g. Determine the percentage of cans con-

(a) 3 bulbs will be defective, (b) not more than 3 taining (a) less than 985 g, (b) more than 1030 g,

bulbs will be defective, (c) at least 2 bulbs will be (c) between 985 g and 1030 g. (11)

defective. (13)

3. Some engineering components have a mean length
of 20 mm and a standard deviation of 0.25 mm.
Assume that the data on the lengths of the compo-
nents is normally distributed.

Section 7

Multiple choice questions on Chapters 28–41

All questions have only one correct answer (answers on page 570).

1. A graph of resistance against voltage for an electri- y y
cal circuit is shown in Figure M3.1. The equation 6 6
relating resistance R and voltage V is:
Ϫ4 0 0
(a) R = 1.45 V + 40 (b) R = 0.8 V + 20 4 x Ϫ6 6x
(c) R = 1.45 V + 20 (d) R = 1.25 V + 20

160 Ϫ6 Ϫ6
145 (ii)
140 (i)
120 y
100 y 6
6
80
Resistance R 70 Ϫ6 0 6 x Ϫ6 0 2 6x
60
40 Ϫ6 Ϫ6

20 (iii) (iv)

0 20 40 60 80 100 120 Figure M3.3
Voltage V
5. A pie diagram is shown in Figure M3.4 where P,
Figure M3.1 Q, R and S represent the salaries of four employees
of a firm. P earns £24 000 p.a. Employee S earns:
5
2. j6 is equivalent to: (a) £40 000 (b) £36 000
(c) £20 000 (d) £24 000
(a) j5 (b) −5 (c) −j5 (d) 5
SP
3. Two voltage phasors are shown in Figure M3.2. If 72°
V1 = 40 volts and V2 = 100 volts, the resultant (i.e.
length OA) is: 108° 120°
(a) 131.4 volts at 32.55◦ to V1 R
(b) 105.0 volts at 32.55◦ to V1
(c) 131.4 volts at 68.30◦ to V1 Q
(d) 105.0 volts at 42.31◦ to V1
Figure M3.4
4. Which of the straight lines shown in Figure M3.3
has the equation y + 4 = 2x? 6. A force of 4 N is inclined at an angle of 45◦ to a
(a) (i) (b) (ii) (c) (iii) (d) (iv)
second force of 7 N, both forces acting at a point,
A
as shown in Figure M3.5. The magnitude of the
V2 ϭ 100 volts
resultant of these two forces and the direction of

the resultant with respect to the 7 N force is:

(a) 3 N at 45◦ (b) 5 N at 146◦
(c) 11 N at 135◦ (d) 10.2 N at 16◦

4N

45°

0B 45° 7N
V1 ϭ 40 volts Figure M3.5

Figure M3.2

376 Engineering Mathematics

Questions 7 to 10 relate to the following 14. The curve obtained by joining the co-ordinates of
information:
The capacitance (in pF) of 6 capacitors is as cumulative frequency against upper class boundary
follows: {5, 6, 8, 5, 10, 2}
values is called;
7. The median value is:
(a) 36 pF (b) 6 pF (c) 5.5 pF (d) 5 pF (a) a historgram (b) a frequency polygon

8. The modal value is: (c) a tally diagram (d) an ogive
(a) 36 pF (b) 6 pF (c) 5.5 pF (d) 5 pF
15. A graph relating effort E (plotted vertically) against
9. The mean value is:
(a) 36 pF (b) 6 pF (c) 5.5 pF (d) 5 pF load L (plotted horizontally) for a set of pul-

10. The standard deviation is: leys is given by L + 30 = 6E. The gradient of the
(a) 2.66 pF (b) 2.52 pF
(c) 2.45 pF (d) 6.33 pF graph is: (c) 6 1
1 (d)
11. A graph of y against x, two engineering quantities,
produces a straight line. (a) (b) 5 5
A table of values is shown below: 6

Questions 16 to 19 relate to the following
information:
x and y are two related engineering variables and p
and q are constants.
For the law y − p = q to be verified it is necessary

x
to plot a graph of the variables.

x 2 −1 p 16. On the vertical axis is plotted:
(a) y (b) p (c) q (d) x
y9 35
17. On the horizontal axis is plotted:
The value of p is: q1
(a) − 1 (b) −2 (c) 3 (d) 0
(a) x (b) (c) (d) p
2 xx
Questions 12 and 13 relate to the following infor-
mation. The voltage phasors V1 and V2 are shown 18. The gradient of the graph is:
in Figure M3.6. (a) y (b) p (c) q (d) x

Section 7 V1 ϭ 15 V 19. The vertical axis intercept is:
(a) y (b) p (c) q (d) x
30°
Questions 20 to 22 relate to the following
V2 ϭ 25 V information:
A box contains 35 brass washers, 40 steel washers
Figure M3.6 and 25 aluminium washers. 3 washers are drawn at
12. The resultant V1 +V2 is given by: random from the box without replacement.

(a) 38.72 V at −19◦ to V1 20. The probability that all three are steel
(b) 14.16 V at 62◦ to V1 washers is:
(c) 38.72 V at 161◦ to V1 (a) 0.0611 (b) 1.200 (c) 0.0640 (d) 1.182
(d) 14.16 V at 118◦ to V1
13. The resultant V1 −V2 is given by: 21. The probability that there are no aluminium
(a) 38.72 V at −19◦ to V1 washers is:
(b) 14.16 V at 62◦ to V1 (a) 2.250 (b) 0.418 (c) 0.014 (d) 0.422
(c) 38.72 V at 161◦ to V1
(d) 14.16 V at 118◦ to V1 22. The probability that there are two brass washers
and either a steel or an aluminium washer is:
(a) 0.071 (b) 0.687 (c) 0.239 (d) 0.343

23. (−4 − j3) in polar form is :

(a) 5∠ − 143.13◦ (b) 5∠126.87◦

(c) 5∠143.13◦ (d) 5∠ − 126.87◦

Multiple choice questions on Chapters 28–41 377

24. The magnitude of the resultant of velocities of 3 m/s y
at 20◦ and 7 m/s at 120◦ when acting simultane-
15
ously at a point is: 10

(a) 21 m/s (b) 10 m/s 5

(c) 7.12 m/s (d) 4 m/s Ϫ2 Ϫ1 0 1 2 3 x
Ϫ5
25. Here are four equations in x and y. When x is
plotted against y, in each case a straight line Ϫ10
results. Ϫ15
Ϫ20
(i) y + 3 = 3x (ii) y + 3x = 3
(iii) y − 3 = x (iv) y = x + 2 Figure M3.7

22 33 31. In an experiment demonstrating Hooke’s law, the
strain in a copper wire was measured for various
Which of these equations are parallel to each stresses. The results included

other?

(a) (i) and (ii) (b) (i) and (iv)

(c) (ii) and (iii) (d) (ii) and (iv) Stress
(megapascals) 18.24
24.00 39.36

26. The relationship between two related engineering Strain 0.00019 0.00025 0.00041
variables x and y is y − cx = bx2 where b and c
When stress is plotted vertically against strain
are constants. To produce a straight line graph it is horizontally a straight line graph results.
Young’s modulus of elasticity for copper, which is
necessary to plot: given by the gradient of the graph, is:

(a) x vertically against y horizontally (a) 96 × 109 Pa (b) 1.04 × 10−11 Pa
(b) y vertically against x2 horizontally (c) 96 Pa (d) 96 000 Pa

y
(c) vertically against x horizontally

x
(d) y vertically against x horizontally

27. The number of faults occurring on a production line Questions 32 and 33 relate to the following Section 7
in a 9-week period are as shown: information:
32 29 27 26 29 39 33 29 37 The frequency distribution for the values of resis-
The third quartile value is: tance in ohms of 40 transistors is as follows:
(a) 29 (b) 35 (c) 31 (d) 28
15.5–15.9 3 16.0–16.4 10
28. (1 + j)4 is equivalent to: 16.5–16.9 13 17.0–17.4 8
(a) 4 (b) −j4 (c) j4 (d) −4 17.5–17.9 6

32. The mean value of the resistance is:

(a) 16.75 (b) 1.0

29. 2% of the components produced by a manufacturer (c) 15.85 (d) 16.95
are defective. Using the Poisson distribution the
percentage probability that more than two will be 33. The standard deviation is:
defective in a sample of 100 components is:
(a) 0.335 (b) 0.251

(c) 0.682 (d) 0.579

(a) 13.5% (b) 32.3% 34. The depict a set of values from 0.05 to 275, the
(c) 27.1% (d) 59.4% minimum number of cycles required on logarithmic
graph paper is:
30. The equation of the graph shown in Fig- (a) 2 (b) 3 (c) 4 (d) 5

ure M3.7 is: (b) 4x2 − 4x − 15 = 0 35. A manufacturer estimates that 4% of components
15 (d) 4x2 + 4x − 15 = 0 produced are defective. Using the binomial distri-
bution, the percentage probability that less than two
(a) x(x + 1) =
4

(c) x2 − 4x − 5 = 0

378 Engineering Mathematics

components will be defective in a sample of 10 4N

components is:

(a) 0.40% (b) 5.19%

(c) 0.63% (d) 99.4%

Questions 36 to 39 relate to the following 60° 3N
30°
information.
2N
A straight line graph is plotted for the equation
y = axn, where y and x are the variables and a and
n are constants.

36. On the vertical axis is plotted: Figure M3.8
(a) y (b) x (c) ln y (d) a

37. On the horizontal axis is plotted: 44. 2∠ π + 3∠ π in polar form is:
(a) ln x (b) x (c) xn (d) a 36

38. The gradient of the graph is given by: (a) 5∠ π (b) 4.84∠0.84
(a) y (b) a (c) x (d) n 2

(c) 6∠0.55 (d) 4.84∠0.73

39. The vertical axis intercept is given by: Questions 45 and 46 relate to the following
(a) n (b) ln a (c) x (d) ln y
information.

Two alternating voltages are given by:

Questions 40 to 42 relate to the following v1 = 2 sin ωt and v2 = 3 sin π volts.
ωt +
information.
4
The probability of a component failing in one year
1 45. Which of the phasor diagrams shown in
Figure M3.9 represents vR = v1 + v2?
due to excessive temperature is , due to exces-
16 (a) (i) (b) (ii) (c) (iii) (d) (iv)

Section 7 1 v2 vR vR v2
sive vibration is and due to excessive humidity

20
1
is .
40

40. The probability that a component fails due to
excessive temperature and excessive vibration is:

285 19 1 v1 v1
(a) (b) (c) (d) (i) (ii)

320 320 80 800 v1

41. The probability that a component fails due to v1
excessive vibration or excessive humidity is:

(a) 0.00125 (b) 0.00257 v2 vR vR v2
(c) 0.0750 (d) 0.1125 (iv)

42. The probability that a component will not fail (iii)
because of both excessive temperature and exces-
sive humidity is: Figure M3.9

(a) 0.914 (b) 1.913 46. Which of the phasor diagrams shown represents
(c) 0.00156 (d) 0.0875 vR = v1 − v2?
(a) (i) (b) (ii) (c) (iii) (d) (iv)
43. Three forces of 2 N, 3 N and 4 N act as shown
in Figure M3.8. The magnitude of the resultant 47. The two square roots of (−3 + j4) are:
force is: (a) ±(1 + j2) (b) ±(0.71 + j2.12)
(c) ±(1 − j2) (d) ±(0.71 − j2.12)
(a) 8.08 N (b) 9 N (c) 7.17 N (d) 1 N

Multiple choice questions on Chapters 28–41 379

Questions 48 and 49 relate to the following 51. The number of people likely to have heights of
information. between 154 cm and 186 cm is:
A set of measurements (in mm) is as follows:
{4, 5, 2, 11, 7, 6, 5, 1, 5, 8, 12, 6} (a) 390 (b) 380 (c) 190 (d) 185

48. The median is: 52. The number of people likely to have heights less
(a) 6 mm (b) 5 mm (c) 72 mm (d) 5.5 mm than 162 cm is:

49. The mean is: (a) 133 (b) 380 (c) 67 (d) 185
(a) 6 mm (b) 5 mm (c) 72 mm (d) 5.5 mm
53. The number of people likely to have a height of
50. The graph of y = 2 tan 3θ is: more than 186 cm is:
(a) a continuous, periodic, even function
(b) a discontinuous, non-periodic, odd function (a) 10 (b) 67 (c) 137 (d) 20
(c) a discontinuous, periodic, odd function
(d) a continuous, non-periodic, even function 54. [2∠30◦]4 in Cartesian form is:

Questions 51 to 53 relate to the following (a) (0.50 + j0.06) (b) (−8 + j13.86)
information.
The mean height of 400 people is 170 cm and (c) (−4 + j6.93) (d) (13.86 + j8)
the standard deviation is 8 cm. Assume a normal
distribution. (See Table 41.1 on pages 367/368)

Section 7

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Section 8

Differential Calculus

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Chapter 42

Introduction to
differentiation

42.1 Introduction to calculus f (3) = 4(3)2 − 3(3) + 2
= 36 − 9 + 2 = 29
Calculus is a branch of mathematics involving or lead-
ing to calculations dealing with continuously varying f ( − 1) = 4(−1)2 − 3(−1) + 2
functions. =4+3+2=9
Calculus is a subject that falls into two parts:
f (3) − f ( − 1) = 29 − 9 = 20
(i) differential calculus (or differentiation) and
(ii) integral calculus (or integration). Problem 2. Given that f (x) = 5x2 + x − 7
determine:
Differentiation is used in calculations involving velocity
and acceleration, rates of change and maximum and (i) f (2) ÷ f (1) (iii) f (3 + a) − f (3)
minimum values of curves. (ii) f (3 + a) f (3 + a) − f (3)

42.2 Functional notation (iv)
a
In an equation such as y = 3x2 + 2x − 5, y is said to be
a function of x and may be written as y = f (x). f (x) = 5x2 + x − 7

An equation written in the form f (x) = 3x2 + 2x − 5 (i) f (2) = 5(2)2 + 2 − 7 = 15
is termed functional notation. The value of f (x)
when x = 0 is denoted by f (0), and the value of f (x) f (1) = 5(1)2 + 1 − 7 = −1
when x = 2 is denoted by f (2) and so on. Thus when
f (x) = 3x2 + 2x − 5, then f (2) ÷ f (1) = 15 = −15
−1
f (0) = 3(0)2 + 2(0) − 5 = −5
and f (2) = 3(2)2 + 2(2) − 5 = 11 and so on. (ii) f (3 + a) = 5(3 + a)2 + (3 + a) − 7

Problem 1. If f (x) = 4x2 − 3x + 2 find: = 5(9 + 6a + a2) + (3 + a) − 7
f (0), f (3), f (−1) and f (3) − f (−1)
= 45 + 30a + 5a2 + 3 + a − 7
f (x) = 4x2 − 3x + 2
f (0) = 4(0)2 − 3(0) + 2 = 2 = 41 + 31a + 5a2

(iii) f (3) = 5(3)2 + 3 − 7 = 41

f (3 + a) − f (3) = (41 + 31a + 5a2) − (41)

= 31a + 5a2

(iv) f (3 + a) − f (3) = 31a + 5a2 = 31 + 5a
aa